# B General Questions about Special Relativity

1. Jul 13, 2016

### NoahsArk

There are a few fundamental questions I wanted to ask about related to special relativity:

1) Firstly, is there an intuitive explanation for length contraction and why lengths are relative? For example, the fact motion is relative is intuitive. E.g. someone sitting inside a train moving 60 miles per hour bouncing a ping pong ball up and down on a table will see the ball moving zero distance in the X direction, whereas someone on the ground will have seen it move in the X direction at 60mph. With length contraction, however, the person on the train measures objects themselves which are outside the train as smaller lengthwise than a person on the ground would measure the same objects. Is there any kind of physical or visual explanation of this like in the case of the ping pong ball? The only explanation that I read which makes sense is a mathematical one: that is, for the observer on the ground velocity = distance/time (v=d/t), and for the one in the train v = d1/t1. Since we know t and t1 are different, we can conclude that d and d1 are different.

Another question is, since time dilation and length contraction are two sides of the same coin, is it possible to say which one is "causing" the other?

2) Regarding the Lorentz transformations the formula for converting distance in one frame to distance in another frame makes much more sense to me (even without doing the lengthy derivation) than the formula for converting time. The formula for distance is X = γX1 + γVT1. That makes sense because to get the distance of an event in a stationary frame from a distance in a moving frame, like a rocket frame, we'd need to know how far the rocket traveled plus how far the event is from the rocket (and multiply both sides of the "+" sign by γ. The formula for converting time, T = γV/C2X1 + γT1, is confusing for me. If an event happened inside the rocket, the X distance would be zero and the formula for converting time would just be T = γT1. Is there an intuitive explanation behind the γV/C2X1 part? If the person in the rocket measures the time of some event happening outside of the rocket, say on a second rocket moving faster than him and even faster than the person on earth, wouldn't the Lorentz transformation for time need to take into consideration how fast the second rocket is moving with respect to the first?

Thanks

2. Jul 13, 2016

### Simon Bridge

1. Depends what you find intuitive. Its basically just perspective... you are used to distant lengths being contracted, now you know that fast lengths are contracted too. Re. Time vs length: it makes no sence to consider one to cause the other just as it makes no sense to say that two sides of a triangle cause the third one.
2. Its time dilation... see answer to 1.

3. Jul 13, 2016

### Staff: Mentor

Back away from trying to intuitively understand length contraction and time dilation, work on nailing down your understanding of relativity of simultaneity instead. Once you have relativity of simultaneity down, you'll find that time dilation and length contraction follow naturally.

The length of an object is pretty obviously the distance between where its two ends are at the same time, so observers with different notions of simultaneity will naturally find different lengths.

Thus, although time dilation and length contraction are two sides of the same coin, you could reasonably say that the coin itself is relativity of simultaneity.

4. Jul 14, 2016

### NoahsArk

Thank you for the responses.

"it makes no sense to consider one to cause the other just as it makes no sense to say that two sides of a triangle cause the third one"

That's an interesting way of looking at it.

"Thus, although time dilation and length contraction are two sides of the same coin, you could reasonably say that the coin itself is relativity of simultaneity."

Nuqatory, I will follow your suggestion and begin studying relativity of simultaneity to try and get more of a foundation. The examples that I've seen so far to explain the concept involve someone in a rocket shooting two beams of light in opposite directions, one to the tail end of the ship and one to the front of the ship. The person in the rocket sees the beams hit both ends simultaneously while someone on earth sees the beam hit the back end first. How does that apply to length contraction though? You said that length has to do with the distance between two places in space at the same time. However, when someone is moving, say in a train, they see the road outside contracting in length. Two ends of a piece of road aren't similar to two beams of light shooting in opposite directions. The end point of a given stretch of road, unlike the light beams, are stationary with respect to one another.

5. Jul 14, 2016

### Staff: Mentor

Let's put a splash of paint on each end of the piece of road, just so that we can all agree about which piece of road we're talking about.

Now, it's easy for a person at rest relative to the road to measure the distance between the two paint marks because the marks aren't moving around - just grab a meter stick, set one end of it on one mark, put a finger at the other end, pick up the meter stick and lay it down again with the end lined up against the finger, pick up the finger and put it at the other end, pcik up the meter stick and place it against the finger, repeat until we reach the other paint mark. You'll notice that we aren't measuring the distance between the positions of the two paint marks at the same time, but that's OK because they aren't moving so the position of the far paint mark when we get to it is the same as when we started the process at the near paint mark. We haven't measured the distance between the two paint marks at the same time, but we've used a procedure that gives us the right answer anyway, so we're happy.

Clearly there's no way of applying the same procedure if we're in the moving train. However, there is a procedure that will work: We ask two lab assistants to stand on the road, one at each of our paint splashes. They're both carrying loaded paintbrushes, and we have directed them to reach out and put a splash of paint on the train as it passes. If they both make their marks on the train at the same time, then we will have two marks on the train that tell us where the endpoints of our piece of road were at the same time - and we on the train have no problem using our meter stick to measure the distance between those two marks because they're not moving relative to us.

But now you'll see where relativity of simultaneity comes in. They have to make their marks at the same time. If we use the train observer's definition of simultaneity, the marks on the train will be closer than if we use the road observer's definition of simultaneity. This fact can be interpreted equally reasonably as "the stretch of road is length-contracted when viewed from the train" or as "the train is length-contracted when viewed from the road".

6. Jul 14, 2016

### NoahsArk

Ok I'll have to think about that one. I don't quite see why road observer and the train observer will see the two men with paint brushes making their splashes at different times. If the train stopped and the person on the ground walked up to the train and measured the distance between splashes, shouldn't someone on the train get the same measurement?

I just started reading the Electrodynamics of Moving Bodies where Einstein gives the first example of the relativity of simultaneity. He uses the example of a rod lying flat on the ground which then begins to move in the X direction. There are also two clocks, A and B, on both ends of the rod being measured. He says the length of the moving rod from the stationary perspective divided by C-V = the time it takes a light beam to travel from clock A to clock B. Then he says the length of the moving rod divided by C + V = the time it takes for the beam to travel back from clock B to clock A. It makes sense that the stationary observer should see it take less time to go back since when it goes back the A clock is moving toward the light. He then goes on to say that this means the clocks are synchronized from the stationary perspective but not from the moving perspective. Isn't it the opposite? As the paper earlier states, two clocks A and B are synchronized when it takes the same amount of time for light to travel from A to B as it does for light to travel from B to A.

7. Jul 15, 2016

### Vitro

@NoahsArk, let's try an exercise. Assume the two clocks A and B are mutually at rest and separated by an unknown distance D. Clock A sends a light pulse when tA = 0 seconds and clock B receives it at tB = 2 seconds. What is the distance between clocks A and B (in their rest frame)?

8. Jul 15, 2016

### NoahsArk

Two light seconds?

9. Jul 15, 2016

### NoahsArk

From what I understand, two clocks, A and B, are synchronized when we set the clocks taking into consideration the time it takes for light to travel from B to A. So, if it takes one hour for light to travel, we'd have to set B's clock an hour ahead. To be more precise, this would mean that the clocks are synchronized from A's perspective. If A's clock is set at 12 and B's clock is set at 1, then, when A's clock strikes 1, he will also see B's clock reading 1 O'Clock since the light will just be arriving from B's location to A. B, however, will see his clocks 2 hours ahead of A's. When his clock reads 2 O'clock, he will see A's clock at 12 since it took the light one hour to transmit A's 12 O'Clock time, and in that hour B's clock, which was initially set for 1, changed to 2.

The definition in Electrondynamics of synchronized clocks, i.e. that two clocks are synchronized when the time it takes for light to travel from A to B = the time it takes for light to travel form B to A, doesn't seem to hold. In the example above, the clocks would be synchronized from A's perspective regardless of the time it took for light to go from A to B- so long as we set the clocks at 12 and 1 (or any other pair of times one hour apart) when it takes light an hour to go from B to A.

Also, the example in Electrodynamics contradicts its own definition of synchronization. It gives two clocks A and B which are moving in uniform motion with respect to one another at some velocity V with respect to a stationary observer. In that example, from the stationary perspective, it will take light longer to go from A to B then it will take light to go the other way.

10. Jul 15, 2016

### Vitro

Hmm, what implicit assumptions did you make to get that answer?

Here's the second part of the exercise I left out previously: at tB = 2 seconds clock B reflects the light pulse back and clock A receives it at tA = 2 seconds. How can that be? What's the revised distance now?

11. Jul 15, 2016

### NoahsArk

I see. The assumption I made was that when TA = 0, TB also = 0.

Regarding your second question, it would mean that the distance is one light second I think.

12. Jul 15, 2016

### Vitro

Yes, in order to measure a one-way distance using light and two clocks the clocks must be synchronized.

In the rest frame of the clocks the distance from A to B is the same as the distance from B to A, and the speed of light from A to B is also the same as from B to A, so if you measure the travel time of light between them as tA-tB (or tB-tA) it must be the same both ways, then the two clocks are synchronized.

In the previous exercise the travel time measured with the two clocks was 2 seconds form A to B and 0 seconds from B to A which means the clocks are out of sync, clock B is 1 second ahead. We can still determine the distance as 1 ls by using just clock A and the round-trip travel time of 2 s.

Note that "when tA = 0, tB also = 0" can only happen in a single frame of reference (simultaneity is relative), which doesn't necessarily have to be the rest frame of the clocks. In the moving rod scenario they happen to be synchronized in the "stationary" frame where they are moving but not because they measure the same time from A to B as from B to A which they obviously do not (that definition only applies in the rest frame of the clocks), it's because they would measure the same time interval over the same distance in both directions, or you can also say that they simultaneously show the same time.

13. Jul 15, 2016

### Staff: Mentor

Yes. In that case everyone will measure the same distance between the marks on the train (and it will be the same distance that the train guy measures while the train is moving - either way we're measuring a train that is at rest relative to us). However, that distance will not be equal to the distance between the two marks on the road after we've used either of the two procedures I describe below.

We could tell them either:
1. Both of you make your mark on the train at the same time, namely when the roadside clock reads noon; or
2. Both of you make your mark on the train at the same time, namely when the train clock reads noon. (This actually requires a bit of cleverness. We can put a whole bunch of clocks on the train, one in every window, all synchronized to read the same time according to an observer on the train. Now we can tell our brush-wielders to make their mark when the moving clock right in front of their nose reads noon, and the marks will end up being made at the same time according to the train clock).
Because of the relativity of simultaneity, these two procedures will result in the marks ending up at different places on the train. If the train clock under the nose of one of our brush-wielders reads noon at the same time that the roadside clock reads noon, then the clock under the nose of the other brush-wielder will not read noon, so the two sets of directions will cause the second guy to make his mark at two different times.
As I said in my first post in this thread, relativity of simultaneity is the key to making sense of all relativity paradoxes. So stick with that paragraph above until you are clear on how it is that the two procedures end up putting the marks in different places on the train.

#2 is the correct procedure for measuring the length of the piece of road from the train - we find where its endpoints were at the same time and then measure the distance between them. That measurement will show that the road is length-contracted from the point of view of the train - the distance between the two marks on the train, measured by a guy with a meter stick on the train, will be less than the distance between the two marks on the road. (Now you may be thinking that because the train is moving this measurement is somehow illegitimate, and that the "real" measurement is the one that we would get if we stopped the train. If you're thinking that, we can go back and redescribe the whole though experiment as if the train is at rest while the road and everything else is moving backwards - and before you reject that perspective, remember that the road is attached to the surface of the earth which is rotating, moving around the sun at several kilometers per second, and moving along with the sun in a long slow orbit around our galaxy, which also isn't standing still).

#1 will show that the train is length-contracted from the point of view of the road. The distance between the marks as measured by train guy or after we have stopped the train will be greater than the distance between the two road marks.

Last edited: Jul 15, 2016
14. Jul 15, 2016

### PeroK

This is not correct. Clock syncronisation does not depend on the delay introduced by the finite speed of light. Indeed, this is the case with all observations. The time you record an observation must take into account how long the light took to reach you. If you see a clock a mile away and here the chime 5 seconds later, then the chime and the clock reading the hour are simultaneous. As an observer, you would need to take into account the finite time both the light and sound take to reach you. You would not conclude that the clock hands reached the hour and the clock chimed at different times.

More fundamentally, relativity deals with the differences between moving reference frames, rather than moving observers. One way to think of a reference frame (which is very useful when studing SR) is a whole series of observers, all at rest with respect to each other and all with synchronised watches. Anywhere an event takes place, there will be an observer on the spot to record the time of the event. After an experiment, all the observers can get together and compare notes and put together a full picture of what happened, where and when in their reference frame. One thing you shouldn't do is include the delay for light to travel from an event to an observer and say that the event happened later than it did: that has nothing to do with SR.

The 1905 paper does not contradict itself. Thinking that will only hinder your attempt to understand it.

15. Jul 15, 2016

### Staff: Mentor

I would recommend learning relativity from a more modern source. Einstein's original paper is a must-read, but probably not the best place to start. The language, notation, pedagogy, and concepts evolved over time.

16. Jul 15, 2016

### Staff: Mentor

What he said.
"Spacetime Physics" by Taylor and Wheeler is one of several possible good choices.

17. Jul 15, 2016

### NoahsArk

"I would recommend learning relativity from a more modern source."
"Spacetime Physics" by Taylor and Wheeler is one of several possible good choices."

Thanks- I was planning to get that book. There was also a good online coursera class on the special theory of relativity which I was taking, but unfortunately it got taken off coursera and is not available anymore.

"One thing you shouldn't do is include the delay for light to travel from an event to an observer and say that the event happened later than it did: that has nothing to do with SR."

Right. If I understand it correctly, someone perceiving two events at different times doesn't mean that those two events did not happen simultaneously in that same observer's frame of reference. E.g. if he is standing in the back of a train and a beam of light from the middle of the train shoots out in both directions, the person in the back will see the light hit the back of the train before the other beam hits the front, but those two events still happened simultaneously for him. Am I understanding it right?

Regarding Electrodynamics, does Einstein use "synchronous" to mean that both clocks tick at the same rate, or that the two clocks are adjusted to the same time? Also, what's the distinction between synchronous and simultaneous?

I must be misreading the paper. He is saying that if two clocks, A and B, are on opposite ends of a moving body, and a beam of light is shot from A to B, and then back from B to A, someone moving with the clocks will see the clocks out of sync, but someone who is stationary in relationship to the clocks will see the clocks in sync. How is it not the other way around? The person who is moving with the clocks will see the beam of light take the same time to go from A to B as it does to go back from B to A. The person who is stationary will see the beam's return from B to A as being faster than it took to get from A to B.

Also, he doesn't explain why the clocks' synchronization should depend on the time it takes light to go from A to B compared with the time it takes to go from B to A. The only explanation I know of is related to the fact that time dilates from the point of view of a stationary observer of clocks on a moving body. If it took longer for the light to go from B to A we could conclude that B was moving relative to A. However, Electrodynamics defines synchronization without even getting into time dilation. Is there a more fundamental reason, then, about why the synchronization of two clocks depends on the time it takes for light to go from one to the other?

18. Jul 15, 2016

### Staff: Mentor

Right. If light hits my eyes at noon, and it was emitted from somewhere two light-minutes away then I know it that the emission happened at 11:58 and therefore happened at the same time ("simultaneously") with everything else that happened at 11:58. That would include something that happened three light minutes away that I don't see until one minute past noon, something that happened right under my nose at 11:58,

The basic issue with relativity of simultaneity is that if two observers are moving relative to one another, events that are simultaneous according to one will not be simultaneous according to the other.
Two clocks are synchronized if they read the same thing at the same time. You're holding one clock in your hand and it reads 11:58; there's another clock two light minutes distant. How would you determine that the two clocks are synchronized? You need to know that the distant clock reads 11:58 as well, and because it's two light-minutes away that means that you'll see it reading 11:58 when your clock reads noon - same as the example above. But this procedure obviously depends on the the light travel time - it won't work if the light doesn't take exactly two minutes to cover the distance.

Last edited: Jul 15, 2016
19. Jul 16, 2016

### Orodruin

Staff Emeritus
No, he will also see them at the same time because the events happened at the same distance from him. Also, you should be very careful when you use the word "see" in SR as it is very confusing to many. What an observer actually sees depends on when the light from an event arrives to it. In order to find this out, you need to account for the travel time of the light, precisely what we are generally not wanting to do when we talk about SR examples. A better nomenclature is to say that the events occur at the same time in his frame.

20. Jul 16, 2016

### NoahsArk

"Two clocks are synchronized if they read the same thing at the same time. You're holding one clock in your hand and it reads 11:58; there's another clock two light minutes distant. How would you determine that the two clocks are synchronized? You need to know that the distant clock reads 11:58 as well, and because it's two light-minutes away that means that you'll see it reading 11:58 when your clock reads noon"

This makes sense and clarifies things more for me. So, one way it seems to me to make sense to define clocks being "synchronized" is: two clocks A and B are synchronized when, after adjusting for the time it took light to travel, the clocks were in the same position at the same time." The examples you used fit that definition. There is still a problem for me, though. Couldn't you have a situation where two unsynchronized clocks, A and B, were two light minutes apart, and they both read 11:58 simultaneously since an observer at clock A, when it was noon time for him, saw clock B's time reading 11:58? If clock B were moving away from clock A, and both happened to read 11:58 at the same time, wouldn't they still be unsynchronized since B is moving away from A?

"But this procedure obviously depends on the the light travel time."

To me it makes sense that to figure out whether or not two events happened simultaneously, you'd need to take into account the light travel time. However, why do we need to take into account the comparative travel time of the light? I.e. the travel time of the light from A to B is vs. the travel time of the light from B to A? In other words, why do we need to know that these two travel times are = to say that the clocks are synchronized?

Also, please let me know if this is correct: if two clocks A and B, which are positioned at the back and front of a rocket, are set into motion when the rocket begins traveling, and both clocks were initially set at noon, a person in the rocket will observe both clocks to be synchronized, but someone on earth, stationary with respect to the rocket, will see the clocks unsynchronized.

21. Jul 16, 2016

### NoahsArk

"Two clocks are synchronized if they read the same thing at the same time. You're holding one clock in your hand and it reads 11:58; there's another clock two light minutes distant. How would you determine that the two clocks are synchronized? You need to know that the distant clock reads 11:58 as well, and because it's two light-minutes away that means that you'll see it reading 11:58 when your clock reads noon"

This makes sense and clarifies things more for me. So, one way it seems to me to make sense to define clocks being "synchronized" is: two clocks A and B are synchronized when, after adjusting for the time it took light to travel, the clocks were in the same position at the same time." The examples you used fit that definition. There is still a problem for me, though. Couldn't you have a situation where two unsynchronized clocks, A and B, were two light minutes apart, and they both read 11:58 simultaneously since an observer at clock A, when it was noon time for him, saw clock B's time reading 11:58? If clock B were moving away from clock A, and both happened to read 11:58 at the same time, wouldn't they still be unsynchronized since B is moving away from A?

"But this procedure obviously depends on the the light travel time."

To me it makes sense that to figure out whether or not two events happened simultaneously, you'd need to take into account the light travel time. However, why do we need to take into account the comparative travel time of the light? I.e. the travel time of the light from A to B is vs. the travel time of the light from B to A? In other words, why do we need to know that these two travel times are = to say that the clocks are synchronized?

Also, please let me know if this is correct: if two clocks A and B, which are positioned at the back and front of a rocket, are set into motion when the rocket begins traveling, and both clocks were initially set at noon, a person in the rocket will observe both clocks to be synchronized, but someone on earth, stationary with respect to the rocket, will see the clocks unsynchronized.

22. Jul 17, 2016

### PeroK

I said this in a previous post: light travel time delaying observations has nothing whatsoever to do with Relativity. It applies equally to classical physics. All observations made over any significant distance need to take into account the signal time. But, this is a complexity of experimental physics that is not core to the theory.

You also need to try to forget about observers constantly looking at each others clocks. Observers use their own local clocks. They don't look at distant clocks. The last paragraph about clocks in a rocket is correct, but let me rephrase it:

If two clocks A and B are synchronised in the rest frame of a rocket, they will not be synchronised in a reference frame in which the rocket is moving.

You could, alternatively, say: someone on Earth would "measure" the clocks to be unsynchronised. But, I urge you again to move away from this concept of what observers actually "see". And start thinking in terms of the rocket's reference frame and the Earth's reference frame.

For example, for the moving rocket I would have two "stationary" observers (with syncronised clocks) positioned so that (in their reference frame) the front of the rocket passes the first observer at the same time as the back of the rocket passes the second. Each records the time this happens on their own clock and observes the clock inside the rocket where they are positioned (this clock is then, for a short time, local to them and they can observe it without any significant light delay).

Afterwards, they compare notes and find that in their reference frame the rocket's clocks were not synchronised.

Doing the experiment this way removes all considerations of light travel time, as only clocks that were local to a given observer were observed.

In any case, understanding reference frames is a critical prerequisite to understanding relativity.

23. Jul 17, 2016

### NoahsArk

"If two clocks A and B are synchronized in the rest frame of a rocket, they will not be synchronized in a reference frame in which the rocket is moving."

Thank you for explaining. I have been doing some research on the relativity of simultaneity. I've found useful info, but most of it is very general, and most of the sites give the same one example. The example the sites give is with two beams of light shooting out from the middle of a train towards each end, and showing that someone on the ground will observe the beam hitting the back first, while someone on the train will see the hit at the same time.

This example makes sense, but I think it would be helpful to have many more examples of other kinds of events that don't involve light beams, and to explain why the simultaneity of those events are relative. Even with the light beams, there are so many other variations of that which could happen. For example, what if two beams each shot out horizontally towards the sides of the train. In that case I don't see why the person on the ground and the person in the train wouldn't agree that both beams hit the walls of the train simultaneously. If instead of a beam of light shooting towards the front and the back, would the gap between the occurrence of the ball hitting the back of the train and it hitting the front be the same gap observed by the person on the ground as the gap he observed between the two beams hitting? I assume that if so, it has something to do with the way velocities add using the Lorentz formula. If there is a book or something on the net that explains in more detail with more examples the relativity of simultaneity, please let me know.

Another question which is interesting for me is: does the relativity of simultaneity mean that if a group of people on a rocket go on a voyage, an observer on earth will judge them to be aging at different rates? Would this also mean that he'd see them all different ages when they came back?

I watched a video using space time diagrams to illustrate relativity of simultaneity:
How can I draw space time diagrams in here to illustrate? I want to learn how to better visualize them. In this video he uses an example of two people, B and B1, moving in uniform motion as person A, who is stationary, observes them. If a beam of light shoots out from the origin of A's system of coordinates, it will go out at a 45 degree angle. B and B1's motion are represented by two lines with the same slope at angles of more than 45 degrees in A's coordinates. Since the beam of light is at a 45 degree angle in A's coordinate system, wouldn't we need to draw a different beam of light for it to be 45 degrees from B's lines of motion? I hope that makes sense since this was not easy for me to describe.

Last edited: Jul 17, 2016
24. Jul 17, 2016

### Mister T

You can create lots of other examples quite easily. Let's say that when the light beam reaches the back of the train Bob is there and he takes a bite out of his sandwich. Alice is at the front of the train and she coughs when the light beam reaches her. So, here's what the example will look like without the light beams:

Alice is at the front of a train when she coughs, Bob is at the back of the train when he takes a bite out of his sandwich. These two events happen at the same time according to observers at rest on the train, but according to observers at rest on the ground Alice coughs after Bob bites.

Of course, without the light beams to explain why there's relative simultaneity, the example is pretty much meaningless!

You are quite right. The spatial separation of the events must have a nonzero component along the line of motion for there to be relative simultaneity.

25. Jul 18, 2016

### David Lewis

No. All that's necessary is for the clocks' hands to be in the same position at the same time. Adjusting for the time it takes light to reach the observer is not necessary.