Understanding Length Contraction and Time Dilation

[DuckAmuck]

Length EXPANSION?!

Okay, I understood length contraction at one point, but now this is looking really wishy washy. Can somebody clear this up for me?


Suppose two reference frames S and S' where S' is moving at v with respect to S. Now, the relativistic transformations are t = \gamma(t' + vx'/c^{2}) and x =\gamma(x' + vt').

Now it is easy to see where time dilation comes from with the first equation: Suppose two events occur at t1' and t2' in the same place x', so one gets the difference of t1-t2 = \gamma(t1' + vx'/c^2 - t2' - vx'/c^2) = \gamma(t1'-t2').

Now if you use the same method with the second equation, but flip position and time such that you are measuring length by measuring the positions of x1 and x2 simultaneously, you get x1 - x2 = \gamma(x1' +vt' - x2' - vt') = \gamma(x1' - x2')

So it looks as though length dilates like time. I know this isn't right, I just don't remember why.
I looked it up on the web, and every site I checked just flips the reference frame for no apparent reason. If THAT is right, couldn't I just do that with time and get time contraction?

 

[cepheid]

"Measuring the positions of x1 and x2 simultaneously" makes no sense. You have to be careful and consider how exactly the measurement is to be done. Furthermore, what is simultaneous in one frame is not, in general, simultaneous in another.

 

[DuckAmuck]

Actually, every book and website I have read says the length of a moving object is measured by finding the positions of the front and back simultaneously (in the S frame, presumably).

 

[cepheid]

Let's say a guy asks his friend to drive by an object at constant speed u so that he can measure its length by noting the time (t1) at which the friend passes the near end and the time (t2) at which he passes the far end, and then take this time interval and multiply it by u to get the length of the object. He asks his friend to also make the measurement by noting the times (t1' and t2') at which the near and far ends of the object pass by him going the other way.

When his friend reports a shorter length than what he measured, he counters by saying, "oh yeah, well I saw that your clock was running slowly, so you recorded a shorter time interval (t2'-t1') than what was actually elapsed. My measurement agrees with what I got using this ruler."

 

[George Jones]

Sorry, I responded too quickly.

Yes, this is confusing. Often, a spacetime diagram, with key events labeled, is the key to solving an SR problem. Then, assign coordinates (for both frames) to the key events, and use either invariance of the interval (preferable, for me) or Lorentz transformations, or both, to solve for unknown quantities.

I have now done this for Lorentz contraction. More to come.

 

[George Jones]

Consider the attached spacetime diagram. The t axis, x = 0, is the worldline of the unprimed observer, and the t' axis, x' = 0, is the worldline of the primed observer.

Conventionally, for time dilation, a clock moving in the unprimed frame and stationary in the primed frame is considered, e.g., the primed observer's watch that follows the worldline of the t' axis. Coordinate time differences between, for example, events O and C are compared for the primed and unprimed frames.

Conventionally, for Lorentz contraction, a rod stationary in the unprimed frame and moving in the primed frame is considered. Consequently, the two ends of the rod both follow vertical worldlines in the attached spacetime diagram. Suppose the left end of the rod follows the t axis and the right end of the rod follows the vertical line ABC. The length of the rod in the unprimed frame is the spatial coordinate difference between events O and A, since these events happen simultaneously (t = 0) in the unprimed frame. The length of the rod in the primed frame is the spatial coordinate difference between events O and B, since these events happen simultaneously (t' = 0) in the primed frame.

Differences between time dilation and length contraction:

1) conventionally (but not necessarily) a clock moving in the unprimed frame is considered for time dilation, while a rod moving in the primed frame is considered for length contraction;

2) one pair of events, O and C, is used for time dilation, while two pairs of events, O and A and O and B, are considered for length contraction.

If anyone wants to see the derivations of time dilation and Lorentz contraction using the attached diagram and invariance of the interval, I'll post them.

 

[Dale]

So it looks as though length dilates like time. I know this isn't right, I just don't remember why.
I looked it up on the web, and every site I checked just flips the reference frame for no apparent reason.

Hi DuckAmuck,

JesseM and neopolitan have a https://www.physicsforums.com/showthread.php?t=299857" on exactly this issue. I don't know if they reached any firm conclusion other than "yeah, it is confusing".

Historically I know that length contraction and time dilation were used to contrast the explanation of the same thing in two different frames. For example, with the usual http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/muon.html#c5" example in the ground frame the muon survives because its half-life time is dilated but in the muon's frame it survives because the distance through the atmosphere is contracted.

 

[neopolitan]

Okay, I understood length contraction at one point, but now this is looking really wishy washy. Can somebody clear this up for me?


Suppose two reference frames S and S' where S' is moving at v with respect to S. Now, the relativistic transformations are t = \gamma(t' + vx'/c^{2}) and x =\gamma(x' + vt').

Now it is easy to see where time dilation comes from with the first equation: Suppose two events occur at t1' and t2' in the same place x', so one gets the difference of t1-t2 = \gamma(t1' + vx'/c^2 - t2' - vx'/c^2) = \gamma(t1'-t2').

Now if you use the same method with the second equation, but flip position and time such that you are measuring length by measuring the positions of x1 and x2 simultaneously, you get x1 - x2 = \gamma(x1' +vt' - x2' - vt') = \gamma(x1' - x2')

So it looks as though length dilates like time. I know this isn't right, I just don't remember why.
I looked it up on the web, and every site I checked just flips the reference frame for no apparent reason. If THAT is right, couldn't I just do that with time and get time contraction?

Length contraction applies to simultaneous events.

Time dilation applies to colocated events.

I drew a lot of spacetime diagrams to try to explain something to JesseM in the thread that Dale mentioned, but I'll try to explain in words.

Think of a rod at rest (with respect to you). It has a length L and, with respect to you, both ends of that rod are simultaneous with you.

If you are in motion with respect to me, with the rod held out in the direction of your motion towards or away from me, then the ends of the rod which are simultaneous to you are not simultaneous to me. At each moment, I see different ends of the rod to what you see. The end of the rod which you are not holding is out of synch with what you see in such a way that it is (was/will be) closer to the end of the rod I see you holding.

We can see that if we use a Lorentz Transformation to find the details for the simultaneous ends of the rod in my frame.

x₀ = 0 is you holding one end of the rod (in your frame)
x₁ = L is the other end of the rod (in your frame)

x₀ = \gamma.(x'₀-vt')
x₁ = \gamma.(x'₁-vt')

L = x₁ - x₀
L' = x'₁ - x'₀ = (x₁ - x₀) / \gamma = L / \gamma

t₀ = \gamma.(t'₀ - vx'₀/c²)
t₁ = \gamma.(t'₁ - vx'₁/c²)

Since these were simultaneous in my frame t'₁ = t'₀ and t'₁ - t'₀ = 0.

\Deltat = t₁ - t₀ = \gamma.(t'₁ - vx'₁/c²) - \gamma.(t'₀ - vx'₀/c²)

\Deltat = \gamma.(-vx'₁/c²) + vx'₀/c²) = -L.v/c².\gamma

So the amount to which the ends of the rod which are simultaneous to me are not simultaneous to you is a function of the speed and the rest length of the rod.

Looking at time dilation: as well as a rod, you have a clock and I have a clock too. I want to know what the time is in my frame between two events which are colocated in your frame.

The problem (confusion) we will run into is that we now have to partially change our priming notation to get the standard time dilation equation (in the Lorentz Transformation).

x₀ is the location of your clock relative to you for the first event and x₁ is the location of your clock relative to you for the second event and since you hold your clock close to you the whole time x₀ = x₁ = 0.

t₀ is the time of the first event according to you and t₁ is the time of the second event according to you so \Deltat = t₁ - t₀ is the time between them. Finding out \Deltat', we have to use a different perspective, and hence different notation:

t'₀ = \gamma.(t₀ - vx₀/c²) = \gamma.t₀
t'₁ = \gamma.(t₁ - vx₁/c²) = \gamma.t₁

\Deltat' = \gamma.\Deltat

or as normally written:

t' = \gamma.t

Personally, I find this fiddling around with the priming notation very confusing and I think it confuses other people. It helps, I suppose, if you note that

t' = \gamma . something

in both instances, but the spatial equations have x' in one instance and L' in the other. For my money, it doesn't help enough, but for others it might suffice

cheers,

neopolitan

 

[JesseM]

Length contraction applies to simultaneous events.

Time dilation applies to colocated events.

I think that needs some elaboration. With length contraction, you aren't talking about a single pair of events, but rather comparing the distance between two different pairs of events that occur on either end of the rod, the first pair being simultaneous in the rod's rest frame and the other pair being simultaneous in the frame where the rod is in motion. With time dilation, on the other hand, you are talking about the time interval between a single pair of events in two different frames, the two events being events on the worldline of the clock you're dealing with. If we had a pair of events that were simultaneous in the unprimed frame and had a separation of dx in that frame, and we wanted to know the separation dx' between the same pair of events in a different frame, then the equation would be:

dx' = \frac{dx}{\sqrt{1 - v^2/c^2}}

So, here dx' would be larger than dx, but dx' does not represent the "length" of anything in the unprimed frame, because it's the distance between events which are not simultaneous in the unprimed frame, whereas "length" always refers to the distance between events which are simultaneous in whatever frame you're using (like simultaneous events on either end of a rod).

Think of a rod at rest (with respect to you). It has a length L and, with respect to you, both ends of that rod are simultaneous with you.

Simultaneity only applies to events, not objects with continuous worldlines like ends of a rod. If two firecrackers are set off at either end of the rod simultaneously in the rod's rest frame, then the distance between these events is equal to the rod's proper length. On the other hand, if two different firecrackers are set off at either end of the rod simultaneously in the frame of an observer who sees the rod in motion, then the distance between these events is equal to the rod's length in this observer's rest frame.

The problem (confusion) we will run into is that we now have to partially change our priming notation to get the standard time dilation equation (in the Lorentz Transformation).

Why do you say the notation is changed? For length, unprimed refers to the length of the rod in the rod's rest frame (and primed refers to the length of the same rod in the frame where the rod is moving); for time, unprimed refers to the time between events on the clock's worldline in the clock's rest frame (and primed refers to the time between these same events in the frame where the clock is moving). So in both cases unprimed refers to measurements made in the frame where the instrument is at rest, and this gives the equations:

dt' = \frac{dt}{\sqrt{1 - v^2/c^2}}

L' = L \sqrt{1 - v^2/c^2}

 

[Saw]

So in both cases unprimed refers to measurements made in the frame where the instrument is at rest, and this gives the equations:

dt' = \frac{dt}{\sqrt{1 - v^2/c^2}}

L' = L \sqrt{1 - v^2/c^2}

Not disagreeing, but just trying to check my understanding:

This means that the clock that measures dt (proper time) does not belong to, i.e., is not at rest in the same frame as the rod whose length is dx (rest length). If we have a blue and a red frame, the equations would look as follows:

dt' = dt/sqrt(1 - v^2/c^2)

L' = L *sqrt(1 - v^2/c^2)

Right?

 

[JesseM]

Not disagreeing, but just trying to check my understanding:

This means that the clock that measures dt (proper time) does not belong to, i.e., is not at rest in the same frame as the rod whose length is dx (rest length). If we have a blue and a red frame, the equations would look as follows:

dt' = dt/sqrt(1 - v^2/c^2)

L' = L *sqrt(1 - v^2/c^2)

Right?

As long as you undestand that unprimed refers to the rest frame of each instrument (rod and clock), and primed refers to the frame where they are in motion, I don't see why it would matter whether the two instruments share the same frame or not. If you consider a clock that is attached to a rod (so they share the same rest frame), and L refers to the length of the rod in their mutual rest frame while dt refers to the time between two events on the clock's worldline in this same frame, then labeling the rod/clock rest frame as red and the frame of an observer who sees them in motion as blue we have the equations:

dt' = dt/sqrt(1 - v^2/c^2)

L' = L *sqrt(1 - v^2/c^2)

This is not to say your own way of writing it was wrong, just that your way of writing it would apply to a case where the clock is at rest in the blue frame while the rod is at rest in the red frame. It just depends on the physical situation you want to consider.