Length contraction and direction of tavel

[Amr Elsayed]

Hi all, I want to make sure of a particular information. Length is contracted only in direction of motion. If I am on a spacecraft moving with high speed, I shall see the universe is contracted just in front of me that I am going to, but there would be no contraction If I look to regions that I'm getting further from ." those regions behind me" ??

 

[Dale]

There would be contraction behind you also because behind is still parallel to the direction of travel. There would be no contraction to the left or right because those are perpendicular to the direction of travel.

 

[Amr Elsayed]

But if it's the same contraction ahead and behind, how can C stay the same? For instance 2 beams of light are traveling to me from front and behind. Time dilation will not be enough to see C is fixed

 

[Dale]

Correct. Time dilation is not enough. You need length contraction and the relativity of simultaneity also. The best way is to use the full Lorentz transform.

 

[Amr Elsayed]

I mean for those 2 beams of light, distances traveled by them are different since I'm moving toward one of them and away from another one. I then should see the one I'm going to is faster. but length contraction will not change it ,because it's same for both distances traveled by beams. If length contracts from front only I shall experience same C from front and behind.
Thanks for response

 

[phinds]

you will see one red shifted and one blue shifted but you will see both moving at c

 

[Amr Elsayed]

Well, I am asking how I will see both traveling at c and length contraction is same for front and behind

 

[Dale]

You need the full Lorentz transform. Not just bits and pieces.

Are you familiar with the Lorentz transform?

 

[Amr Elsayed]

Sorry, but Lorentz transformations are time dilation and length contraction, and inertial mass increase is to conserve momentum.

 

[Mentz114]

Sorry, but Lorentz transformations are time dilation and length contraction, and inertial mass increase is to conserve momentum.

The speed of light is invariant under Lorentz transformation, which as you say takes into account those things you mention.

(except for 'inertial mass increase')

 

[Amr Elsayed]

Yes, but my question is about how this purpose " invariant C" can be achieved by Lorentz transformations. I get it if length is contracted only from front. But I don't get how it happens if length is contracted from both front and behind.

 

[Mentz114]

Yes, but my question is about how this purpose " invariant C" can be achieved by Lorentz transformations. I get it if length is contracted only from front. But I don't get how it happens if length is contracted from both front and behind.

If you start with two events $(t_1,x_1)$ and $(t_2,x_x)$ so that $(t_1-t_2)^2-(x_1-x_2)^2=0$, if you transform these events then this relation still holds.

Messing about with 'length contraction' will not lead to better understanding. Use the LT ( as DaleSpam has said).

 

[Nugatory]

Sorry, but Lorentz transformations are time dilation and length contraction, and inertial mass increase is to conserve momentum.

No, none of those things are the Lorentz transforms (although they are consequences of the Lorentz transforms). The Lorentz transforms are the more general equations from which length contraction and time dilation (and a whole bunch of other interesting stuff, such as $E=mc^2$) are derived.

To understand this problem properly, imagine two rods joined end to end, each one meter long as measured in a frame in which they are at rest. You are flying past them at great speed. At the exact moment that you are lined up with the junction between the two rods, one them is sticking out behind you and the other is sticking out in front of you.

Where exactly is the front end of the front-pointing rod at that moment? The distance between that point and where you are is of course the length of the front-pointing rod in the frame in which you are at rest, because you're at the other end of that rod at that moment.

Where exactly is the back end of the back-pointing rod at that moment? The distance between that point and where you are is of course the length of the back-pointing rod in the frame in which you are at rest, because you are at the other end of that rod at that moment.

It takes a bit of algebra with the Lorentz transforms, but you'll find that the two lengths are the same and equally contracted from the one-meter length in their rest frame. An easier way of getting this result is simply to work through the derivation of the length contraction formula starting from the Lorentz transforms; you'll see that the derivation works equally well no matter whether some of the object is behind you or in front of you.

 

[jartsa]

I mean for those 2 beams of light, distances traveled by them are different since I'm moving toward one of them and away from another one. I then should see the one I'm going to is faster. but length contraction will not change it ,because it's same for both distances traveled by beams. If length contracts from front only I shall experience same C from front and behind.
Thanks for response

In our universe you can adjust whether a light beam hits you earlier or later, simply by adjusting your speed. If you consider that a change of the speed of the beam, then the speed of the beam changes.

Physicists mean something else by speed: How many meters something travels in one second.

 

[harrylin]

Hi all, I want to make sure of a particular information. Length is contracted only in direction of motion. If I am on a spacecraft moving with high speed, I shall see the universe is contracted just in front of me that I am going to, but there would be no contraction If I look to regions that I'm getting further from ." those regions behind me" ??

See post #4 by Dalespam. It has to do with setting clocks at a distance, as explained in §1 of http://www.fourmilab.ch/etexts/einstein/specrel/www/ as well as the last sentence of §3 . The effects of length contraction and time dilation are small compared to the effect of setting distant clocks differently.

Personally, the following exercise was most useful for me. Imagine a system that according to you is moving (such as a train, with a guy who is sitting in the train and who assumes that the train is "in rest". Now work out how that guy will set clocks in the train "on time" with the help of light or radio signals. You will find that he will set clocks in front of the train, according to your reckoning, behind on clocks that are in the rear.
As a result of his synchronization he will next "measure" what he first made, which is that the speed of light wrt to the train in forward direction is the same as in backward direction. According to you that is not true, and it's due to his wrong settings of clocks. But he will say the same about you.

Note that this clock setting is not automatic. When the train stops, the clocks will be found to be wrong (or it will be found that according to the clocks, the speed of light is not c in both directions). In that sense it is not so that "c stays the same"

 

[Amr Elsayed]

but you'll find that the two lengths are the same and equally contracted

Physicists mean something else by speed: How many meters something travels in one second.

Personally, the following exercise was most useful for me. Imagine a system that according to you is moving (such as a train, with a guy who is sitting in the train and who assumes that the train is "in rest". Now work out how that guy will set clocks in the train "on time" with the help of light or radio signals. You will find that he will set clocks in front of the train, according to your reckoning, behind on clocks that are in the rear.
As a result of his synchronization he will next "measure" what he first made, which is that the speed of light wrt to the train in forward direction is the same as in backward direction. According to you that is not true, and it's due to his wrong settings of clocks. But he will say the same about you.

I now know why length is contracted from both front and back. But the problem is how I would measure same speed of light.
What I know is because C must be same, there is time dilation and length contraction. I don't get how for distant moving clocks with same velocity there is a change in speed of time flow but what I got from you Harald is he will set those clocks to run for different speeds to measure same C according to them.

But let me explain what exactly I mean. suppose there is a spacecraft moving at 0.9C and at an exact moment it is far 3*10^8 meters from a space station. craft is moving away from station .At that moment the station lights a beam into the space craft. The distance between the craft and the station measured by the moving craft is 1307669683 using length contraction. and since C must be 3*10^8, then time for light beam to reach the craft must be 0.4358898944 second from craft's perspective. This means It should take 1 second from perspective of station according to time dilation, but in deed because the craft is moving away by 0.9C it should take 10 seconds for light beam to reach craft from station's perspective not 1 second. I feel confused about it

 

[Nugatory]

But let me explain what exactly I mean. suppose there is a spacecraft moving at 0.9C and at an exact moment it is far 3*10^8 meters from a space station. craft is moving away from station .At that moment the station lights a beam into the space craft. The distance between the craft and the station measured by the moving craft is 1307669683 using length contraction. and since C must be 3*10^8, then time for light beam to reach the craft must be 0.4358898944 second from craft's perspective. This means It should take 1 second from perspective of station according to time dilation, but in deed because the craft is moving away by 0.9C it should take 10 seconds for light beam to reach craft from station's perspective not 1 second. I feel confused about it

You are still confused because you still trying to use time dilation and length contraction.

You must stop using time dilation and length contraction. You must learn what the Lorentz transformations are - they are not time dilation and length contraction. After you have learned the Lorentz transformations, you can use them to understand this problem.

 

[jartsa]

I now know why length is contracted from both front and back. But the problem is how I would measure same speed of light.

Take two clocks, one meter stick, and two assistants. Set the clocks to the same time. Order the assistants to pick a clock and to go to the ends of the meter stick. Tell them to move slowly to avoid disturbing the clocks by time dilation. Tell the assistants to stop their clock when they see a light pass by. To calculate the speed of the light that passed, divide the distance of the clocks by the time difference of the clocks.

A moving person will disagree with you about these three things:
1: the length of the meter stick
2: how fast the clocks tick
3: whether the assistants disturbed the clocks by time dilation while moving the clocks

 

[Dale]

What I know is because C must be same, there is time dilation and length contraction. I don't get how for distant moving clocks with same velocity there is a change in speed of time flow but what I got from you Harald is he will set those clocks to run for different speeds to measure same C according to them.

The formulas for time dilation and length contraction are simplified special cases of the Lorentz transform. They only apply in special circumstances.

The time dilation formula applies for a clock that is at rest in one frame and moving in the other. Light is not at rest in any frame, so the time dilation formula does not apply.

The length contraction formula applies for two points at rest wit respect to each other which are at rest in one frame and moving in another. Again, this does not apply for light.

The formula you need to use is the Lorentz transform. The simplified formulas just don't apply as-is. With the Lorentz transform we simply start with:
$$x=ct$$ then we substitute in the Lorentz transform to obtain $$\frac{t'v+x'}{\sqrt{1-v^2/c^2}}=\frac{c^2 t'+v x'}{c\sqrt{1-v^2/c^2}}$$ which simplifies to $$x'=ct'$$

 

[harrylin]

[..] what I got from you Harald is he will set those clocks to run for different speeds to measure same C according to them.

Sorry if it was not clear (did you read the clock synchronization procedure?), but those clocks run at the same speed. What you were supposed to get from Einstein is that you set the times of the clocks such that it looks as if the speed of light with respect to the train is the same in both directions.

Did you try to do what I suggested you to do? In order to really understand it, such things should not be said but done, as an exercise! Once more: please sketch for yourself the observer in the moving train with a clock in every wagon, and determine for yourself what happens when he uses Einstein's clock synchronization procedure. You should find that according to you, he sets the clocks at wrong times.

PS this is how it should look like. You draw two lines, one for the train and one for the train station.

Along the lines you indicate a few clocks on the platform, and also directly next to them also in the train (for example one on each end, and one in the middle):

C'1-------------------C'2-----------------C'3_train

C1-------------------C2-----------------C3_station

Below them you put clock readings. For example your three clocks (on the platform in the station) indicate 0, 0 and 0 seconds. Assuming that C'2 also reads 0 at that instant, what will the other clocks in the train read? They will not read 0!

 

[Amr Elsayed]

Light is not at rest in any frame, so the time dilation formula does not apply.

As a moving thing, I determine its velocity by distance over time " with my measures" . I meant to apply length contraction on the distance that light will cover. And time dilation for time the light would need to cover the distance. I then need to understand Lorentz transform first, but do I need calculus or function derivation to understand it ??

Sorry if it was not clear (did you read the clock synchronization procedure?), but those clocks run at the same speed. What you were supposed to get from Einstein is that you set the times of the clocks such that it looks as if the speed of light with respect to the train is the same in both directions.

In deed, it's my English problem again, not because your reply wasn't clear
I have some problems with English of the paper as some expressions are not clear to me even by translation, but what do you mean by wrong times ?? Do you mean different times ? I suppose that you mean the following by the exercise : For a moving train with 2 guys for example with 2 clocks . One of them is in the front of the train, and the other is at the back of the moving train. At a moment when the train is midway between 2 laser projectors " train going toward one and away from the other since the train is moving" . The 2 guys on the train agree to set their clocks on 5 when they receive the laser. For them, both clocks are on time " pointing to same time" since C must have same speed and it was same distance so it will arrive to both guys at same time from their perspective. However, for a third guy on the ground he will see that laser strikes the guy in front before the laser strikes the guy in the back, so their clocks will not be on time " point to same time" from third guy's perspective.
I get it, but here you had an assumption that speed of light is constant for all observers. I know it's a fact proved practically, but the question is how to get same C mathematically which said to need Lorentz transformations.
Another problem is I didn't study calculus which I think it's necessary to understand Lorentz transformations
Sorry for my bad English as It was my problem not to get what you meant from first message

 

[Dale]

I then need to understand Lorentz transform first, but do I need calculus or function derivation to understand it ??

You might need calculus in some derivations, but it only requires basic algebra to use it.

 

[harrylin]

[..] I suppose that you mean the following by the exercise : For a moving train with 2 guys for example with 2 clocks . One of them is in the front of the train, and the other is at the back of the moving train. At a moment when the train is midway between 2 laser projectors " train going toward one and away from the other since the train is moving" . The 2 guys on the train agree to set their clocks on 5 when they receive the laser. For them, both clocks are on time " pointing to same time" since C must have same speed and it was same distance so it will arrive to both guys at same time from their perspective.

Right!

However, for a third guy on the ground he will see that laser strikes the guy in front before the laser strikes the guy in the back, so their clocks will not be on time " point to same time" from third guy's perspective.

Almost right: it's the other way round. Because the front of the train is running away from the light that tries to catch up with the train, it takes extra time for the light to reach the clock in the front, according to you. On the rear it's the opposite: the rear of the train speeds towards the light, and thus it will reach the light earlier.

I get it, but here you had an assumption that speed of light is constant for all observers.

Think of what the guys in the train will "measure" as speed of light, using their clocks after they have done the clock synchronization As the clocks tick at the same rates, they will keep the same time difference. As a result the speed of light (the measured speed of light) is the same in both directions also for them. However, for the speed of light to be exactly the same in all directions and also exactly the same for both reference systems, you need time dilation and length contraction to make it perfect. Time dilation and length contraction are done by nature, and clock synchronization is done by us.

I know it's a fact proved practically, but the question is how to get same C mathematically which said to need Lorentz transformations.

There are several ways to derive it; you can do it in small steps, and that is how it was first done by Lorentz, or you can do it in one go in an elegant way, as next done by Einstein.

Another problem is I didn't study calculus which I think it's necessary to understand Lorentz transformations
Sorry for my bad English as It was my problem not to get what you meant from first message

That's OK. You can try if you can understand most of this derivation:

http://www.bartleby.com/173/a1.html
It links to the drawing here: http://www.bartleby.com/173/11.html

However, it is likely too hard on you... A simplified variant of the way Lorentz derived it may be easier.

 

[Amr Elsayed]

Think of what the guys in the train will "measure" as speed of light, using their clocks after they have done the clock synchronization As the clocks tick at the same rates, they will keep the same time difference. As a result the speed of light (the measured speed of light) is the same in both directions also for them. However, for the speed of light to be exactly the same in all directions and also the exactly the same for both reference systems, you also need time dilation and length contraction. Time dilation and length contraction are done by nature, and clock synchronization is done by us.

I don't get how simultaneity of events has to do sth with time dilation and length contraction to keep C same. Since I want to measure a velocity, I should concentrate on distance and time. However, when I use only them I don't get same C. I was told to understand Lorentz transform first, and I will try to do and read links you attached. But I am not sure if this will make me calculate same C

 

[harrylin]

I don't get how simultaneity of events has to do sth with time dilation and length contraction to keep C same.

You remarked correctly that it is impossible to find c for the speed of light in all directions if we only think of length contraction and time dilation - it is not possible to find the same speed in forward direction as in backward direction without setting the clocks accordingly. Did you try to calculate it in more detail, with realistic numbers? You should find that adapted clock synchronization already works very well to get approximately c in all directions for not too fast moving systems. That was called "local time".

Next, additional corrections were made by assuming length correction and time dilation. Those three things together gave the Lorentz transformations.

Since I want to measure a velocity, I should concentrate on distance and time. However, when I use only them I don't get same C. I was told to understand Lorentz transform first, and I will try to do and read links you attached. But I am not sure if this will make me calculate same C

For distant time you first need to synchronize your clocks. Also the Lorentz transform is based on that fact.

 

[Janus]

I don't get how simultaneity of events has to do sth with time dilation and length contraction to keep C same. Since I want to measure a velocity, I should concentrate on distance and time. However, when I use only them I don't get same C. I was told to understand Lorentz transform first, and I will try to do and read links you attached. But I am not sure if this will make me calculate same C

Consider Einstein's Train example.
You have a train with an observer at the midpoint between the ends. you also have an observer standing along the tracks. Lightning strikes the end of the trains when, according to the track-side observer the train observer is passing him. Thus he sees the light from the strikes at the same time and determines that the strikes occurred simultaneously. Thus, according to the frame of the tracks, events look like this:

Now consider how things look from the frame of the train. The lightning strikes the ends of the train, and each lightning strike has to happen when the respective end of the train is next to the same red dot as it was according to the track frame. The light from each flash must also arrive at the track-side observer at the same time just like in th efirst animation. The light from either flash must also hit the train observer when he is next to the same point of the tracks as it does according to the track frame. (In other words, any event that happens in any frame must happen in the other frame.)

Now here is where you have to take length contraction into account. In the track frame as shown above, the train is moving so it is length contracted. So it is the length contracted train that fits between the two red dots that mark where the lightning strikes occur. In the train frame, the train is its non-contracted proper length, and it is the tracks that are length contracted. Thus the distance between the red dots is shorter than the length of the train and the ends of the train do not align with these dots at the same time. Since the event of the lightning striking an end of the train when it is next to a red dot is common to both frames, this means the lighting strikes cannot occur at the same time in the train frame. And in order for the light from each strike to reach the track observer at the same time and to meet the requirements of all the other common events between frames, the light must expand outward at a constant speed from the points of the strikes. Thus from the frame of the train events look like this.

Time dilation factors in when you try to measure how long it takes for the light to travel from the end of the train to the observer. If you put clocks at the ends of the train that are synchronized in the train frame, they will not be so in the track frame. So in the train frame if you want to measure how fast light is traveling, you note the time on one clock when the light leaves it, then note the time on the clock at the other end of the when it arrives, take the difference in readings and divide this into the length of the train.

You can find the speed of the light in the track frame using the same two clocks, only the method will be different. First you have to account for the length contraction between the clocks, then work out how far the train moves while the light is between clocks. This gives you the total length the light traveled. In order to get the time using these same tow clocks, you have to account for the fact that the clocks are not synchronized in your frame but started with a time offset. Then you factor in time dilation to work out how fast the clocks are running compared to your own. Once you do this, you can determine how much time it took by your clock for the light to pass between the clocks. Divide this into the distance the light traveled in your frame and you will get the same answer for the speed of light as someone in the train does.

 

[Dale]

I don't get how simultaneity of events has to do sth with time dilation and length contraction to keep C same.

Simultaneity is defined in order to keep the one way speed of light equal to c.

You need to use the full Lorentz transform here, not just length contraction and time dilation, as you have been told multiple times now.

 

[Amr Elsayed]

Consider Einstein's Train example.

Did you try to calculate it in more detail, with realistic numbers?

I did understand what you meant Janus, you mean light travels always at same speed and to cover the difference of time between receiving the 2 beams of light, this is because one lighted before the other.However, I did some real calculations, and I got C to be about 275000 kilometers per second. I hope you will tell me where I made a mistake in the following calculations: the same train is moving with 0.9C and it has a proper length of 300000 kilometers. For an observer on the platform he will measure length of train as 13076696.83 meters and the train will have moved 619422481.5 meters in 2.500631499 seconds, so total distance traveled by light from perspective of a still person on the platform is 750189449.8 meters. By dividing I will get exact C because that's how I got that time, but when I use time dilation to get time should be measured by observers on the train" difference in time between the clock in front and the one in the back when light moves between them" I get it as 1.09 seconds instead of 1 second, so observers on train should measure C as 300000000/1.09 which gives about 275000 kilometers per second. I feel like the mistake is about simultaneity, but used clocks should be on time according to observers on the train.
Thanks for caring

as you have been told multiple times now

I know I was told multiple times to search after Lorentz transforms first, but I was not consistent they will not affect it, I just wanted more explanation for how simultaneity will have a role.

 

[Nugatory]

I hope you will tell me where I made a mistake in the following calculations:
the same train is moving with 0.9C and it has a proper length of 300000 kilometers. For an observer on the platform he will measure length of train as 13076696.83 meters and the train will have moved 619422481.5 meters in 2.500631499 seconds, so total distance traveled by light from perspective of a still person on the platform is 750189449.8 meters.

Your mistake is that you are using time dilation and length contraction. Until you stop using time dilation and length contraction you will get confusing and inconsistent results because those formulas do not apply to this problem.

For this "why does the speed of light come out to be $c$?" problem, here's what you do:

First, we pick a frame. For this example, I'll pick the frame in which the platform is at rest because you've already done most of the work for that case, and I'll put the origin of the coordinates at the point in spacetime where the light flash is emitted.

Using this frame, we write down the $(x,t)$ coordinates of the point in spacetime where the light signal is emitted, which we call $x_0$ and $t_0$; $x_0=0$ and $t_0=0$ because of where we put the origin.

Using this frame, write down the $(x,t)$ coordinates of the point in spacetime where the light is received. Call these $(x_1,t_1)$ and because you've done the calculation we know that $x_1=750189449.8$ meters and $t_1=2.500631499$ seconds.

We calculate the speed of light as measured in this frame: The speed is the distance divided by time $(x_1-x_0)/(t_1-t_0)$ and of course is comes out to be $c$.

Now comes the part that you've been missing. To calculate the speed of light in the train frame, we do not use the time dilation and length contraction formulas.

Instead, we need the coordinates of both events in the frame in which the train is at rest (we'll call these $(x'_0,t'_0)$ for the light signal being emitted and $(x'_1,t'_1)$ for the light signal being received). Once we have these, we can calculate the speed of light in the frame at which the train is at rest: the speed is the distance divided by time, $(x'_1-x'_0)/(t'_1-t'_0)$.

The Lorentz transformations tell us how to calculate $x'_0$ and $t'_0$ from $x_0$ and $t_0$ and how to calculate $x'_1$ and $t'_1$ from $x_1$ and $t_1$. Try it... You will get $x'_1=172105263$, $t'_1=0.573684209$, $x'_0=0$ and $t'_0=0$, and the speed of light calculated in the train frame will be $c=3\times{10}^8$ just as it was in the platform frame.

 

[Dale]

but when I use time dilation to get time should be measured by observers on the train"

As I mentioned previously, this is wrong. The time dilation formula does not apply here. The time dilation formula is a special case of the Lorentz transform. It is valid only when the time is measured by a clock at rest in one of the frames. That is not the case here. In this case, measuring the speed of light is done by a pair of synchronized clocks in each frame. And the synchronization procedure guarantees that the speed of light is c.

Physics is about more than taking formulas and filing in numbers. For every equation you must know what the variables mean and when the equation applies or does not apply. I don't know how we can be more clear here. The time dilation formula is insufficient, you need to use the Lorentz transform.

 

[Stephanus]

There would be contraction behind you also because behind is still parallel to the direction of travel. There would be no contraction to the left or right because those are perpendicular to the direction of travel.

Ahhh, I see.

 

[Stephanus]

The speed of light is invariant under Lorentz transformation, which as you say takes into account those things you mention.

(except for 'inertial mass increase')

I'm sorry what do you mean?
There is c' to c as there is t' to t and x' to x?

 

[Stephanus]

But if it's the same contraction ahead and behind, how can C stay the same? For instance 2 beams of light are traveling to me from front and behind. Time dilation will not be enough to see C is fixed

you will see one red shifted and one blue shifted but you will see both moving at c

Okay...
Can I ask a question here?
If I'm with my two friends B and F.
Clocks are synchronized, reset at the same time and preprogrammerd.
B is behind me some trillion KM, and F is in front of me with approximately the same distance as B and me.
The three of us travel together at high speed, and suddenly, because we are preprogrammed before. They B and F emit a beam of light to me.
Will I see B's red shifted and F's blue shifted?

 

[Janus]

Okay...
Can I ask a question here?
If I'm with my two friends B and F.
Clocks are synchronized, reset at the same time and preprogrammerd.
B is behind me some trillion KM, and F is in front of me with approximately the same distance as B and me.
The three of us travel together at high speed, and suddenly, because we are preprogrammed before. They B and F emit a beam of light to me.
Will I see B's red shifted and F's blue shifted?

No, if B and F have no relative velocity with respect to you, you will see no Doppler shift.

 

[Stephanus]

The speed of light is invariant under Lorentz transformation...

I'm sorry what do you mean?
There is c' to c as there is t' to t and x' to x?

Terribly sorry. My poor English (or haste?)

http://dictionary.reference.com/browse/invariant?s=t
adjective
1. unvarying; invariable; constant.
I tought the opposite, wish I can delete my post. Sorry to cause confusing.