# Confusion about time dilation and length contraction

1. Feb 24, 2014

### scinoob

Hi all, I tried searching for this but failed to find an answer to my question. I am having an issue with properly interpreting the equations for time dilation and length contraction. Let's assume that I am standing still and a train is passing by next to me (moving with uniform velocity). Let:

Δx = a length of an object on the train measured by me
Δx' = the length of the same object on the train measured by a person on the train
Δt = a time interval measured by a clock I'm holding in my hand
Δt' = the same time interval measured by a clock held by a person on the train
v = velocity of the train

Then we have the following equations (the Lorentz transformations):

Δx' = $\frac{Δx-vΔt}{\sqrt{1-v^2/c^2}}$ (1)
Δt' = $\frac{Δt-vΔx/c^2}{\sqrt{1-v^2/c^2}}$ (2)

First, let's assume that I measure the length Δx instantaneously so that Δt=0. Then eq. 1 becomes:

Δx' = $\frac{Δx}{\sqrt{1-v^2/c^2}}$

Second, let's assume that in a new experiment I measure a time interval. Since I'm not moving, Δx=0, so the second equation becomes:

Δt' = $\frac{Δt}{\sqrt{1-v^2/c^2}}$

Given that the denominator is some number between zero and one, it turns out that both the length Δx that I measure, as well as the time interval Δt I measure are smaller than those measured by a person on the train (Δx' and Δt', respectively).

Now, I understand the example with the simple "light clock" which demonstrates time dilation. The light goes up and down and every cycle is 1 unit of time, for the person on the train. For me, it takes longer for light to complete a cycle, since it moves diagonally, so if I compared a clock I hold in my hand to the light clock on the train, I would conclude that the clock on the train is running slower than mine. But that's not what the equation above is telling me. It tells me that whatever Δt (the time interval I measured) is, Δt' will be higher. So, if I measured 1 second, the person on the train will measure, say, 2 seconds.

Similarly, if I measure Δx to be 1 meter, the person on the train will say Δx' is 2 meters.

So, it seems like the equations are telling me that time is actually running faster for the person on the train and the lengths are extending, not contracting.

There clearly is some inconsistency in how I interpret the terms, or I'm making some other type of an error. Please help me resolve this issue :)

Last edited: Feb 24, 2014
2. Feb 24, 2014

### A.T.

The clock & object move relative to you, so you see the clock slowed and object shortened.

3. Feb 24, 2014

### tiny-tim

hi scinoob! welcome to pf!
no

the actual length (of an object on the train) is ∆x' (the length measured on the train) …

no matter what the speed of the train, that is what an observer ont eh train will measure​

the contracted length is the length of the same thing measured by anyone else

so ∆x < ∆x'

(similarly for ∆t)

4. Feb 24, 2014

### scinoob

Thanks for the reply, tiny-tim!

I think the confusion comes from something I heard in an online physics course I was watching on Youtube. There the professor (Ramamurti Shankar) said that two observers traveling at a uniform velocity with respect to each other both accuse each other of having "shorter meter sticks" and "slower running clocks". So I am allowed to say that the length measurements the person on the train makes are shorter than they really are. So, if the person on the train says "the object here is 1 meter long", then I can say "no, it's actually 2 meters long".

Judging by your answer now, it seems like the confusion is caused by the following. You're only allowed to accuse the other person on the train of a shorter meter stick if they are trying to measure something that's not on the train with it. Similarly, you accuse them of having a slowly running time if they are trying to measure the time interval between two events which are not happening on the train (i.e., in their reference frame).

But when it comes to measurements they make in their own reference frame, you're saying the opposite? Or something else?

5. Feb 24, 2014

### tiny-tim

hi scinoob!
no

you're making this very complicated

in a way, i prefer A.T.'s answer to mine

if two people both have metre sticks (ie 1 metre long when stationary),

then each measures the other's metre stick as shorter than 1 metre

6. Feb 24, 2014

### scinoob

I think you're right, I am overthinking it. But now it's almost clear :) There is one last issue that I think would finally allow me to close the last page of this chapter.

The Lorentz transformation equations for length and time (equations 1 and 2 from my original post) seem quite symmetric. Yet, another pair of equations (often used for illustrating time dilation and length contraction) aren't. I'm talking about:

L = L_0/γ
T = T_0*γ

Why are we dividing in one case and multiplying in the other? I understand that we have to, in order to actually get length contraction and time dilation, but I don't get how it's conceptually derived from the Lorentz transformation equations. Or rather, I don't get the intuition behind the fact that in the Lorentz equations there is symmetry when it comes to the relationship between (t',x') and (t,x) and dividing/multiplying by the Lorentz factor, whereas in the two equations above there isn't. How do we get one from the other?

7. Feb 24, 2014

### Staff: Mentor

The Lorentz transforms are symmetric in $x$ and $t$, but there's a subtle difference in the definition of the phenomena that we call "length contraction" and "time dilation" that causes an asymmetry in how we write their formulas. Time dilation is the answer to the question "One second has passed in my unprimed stationary frame; how much time is that in the primed moving frame?". Length contraction goes the other way: "The rod is one meter long in the primed moving frame; how long is that in my stationary unprimed frame?". That's why one multiples and the other divides.

However the best way to see the difference is to actually derive the length contraction and time dilation formulas from the Lorentz transforms. Time dilation first: In the unprimed coordinates, my clock is at event (Δt,0) and the moving clock is at event (Δt,vΔt) for some value of Δt. What is the t' coordinate of that second event, as function of Δt?

Length contraction is a bit trickier because you're looking for the ratio of the distance between two pairs of events. If we start with left end of the meter stick at the origin (0,0)in both sets of coordinates:
- The length of the meter stick according to the unprimed stationary observer is the distance between that point and the point where the right-hand end of the meter stick is at time 0 in the unprimed coordinates.
- The length of the meter stick according to the primed moving observer is the distance between that point and the point where the right-hand end of the meter stick is at time 0 in the primed coordinates.
The Lorentz transforms and a bit of algebra will get from there to the length contraction formula.

8. Feb 24, 2014

### Staff: Mentor

First consider a clock which is stationary in reference frame S. In this frame, the time between ticks is $\Delta t$. The clock does not move during this time interval, so $\Delta x = 0$. Plugging into your Lorentz transformation equations we get in reference frame S’:
$$\Delta x^\prime = -\gamma v \Delta t\\ \Delta t^\prime = \gamma \Delta t$$
$\Delta t^\prime$ is the time between ticks in reference frame S’. It’s larger than $\Delta t$ which agrees with our notion of time dilation. Also, in this frame the clock is moving, and $\Delta x^\prime$ is the distance the clock moves between two ticks. So far so good.

Now consider a rod with length L. In general, we determine its length by measuring the positions of the two ends at the same time. If the rod is stationary, we don’t need to measure the ends at the same time, but if the rod is moving, we do; so for consistency, let’s agree to measure the ends at the same time even if the rod is stationary. Then in frame S, $\Delta x = L$ and $\Delta t = 0$. Plugging into the Lorentz transformation, we get in frame S’:
$$\Delta x^\prime = \gamma \Delta x\\ \Delta t^\prime = -\gamma v \Delta x / c^2$$
The first equation looks like it indicates length expansion, not length contraction! For length contraction, we expect to divide by $\gamma$, not multiply.

The problem here is that whereas $\Delta x = L$, the length of the rod in frame S, $\Delta x^\prime \ne L^\prime$, the length in frame S’. The two position measurements that we made at the same time in frame S, do not take place at the same time in frame S’; they’re separated by the time interval $\Delta t^\prime$ given above. During that time, the rod moves, so those position measurements do not give us the length of the rod directly. We have to allow for the distance that the rod moves during that time:
$$L^\prime = \Delta x^\prime + v \Delta t^\prime$$
Plugging in $\Delta x^\prime$ and $\Delta t^\prime$, and doing some algebra, we end up with
$$L^\prime = \Delta x / \gamma = L/\gamma$$
which does indicate length contraction.

9. Feb 24, 2014

### scinoob

Thanks a lot guys, really appreciate it. I think I finally understand it now.

Does Δx' actually have any meaningful interpretation then? With my updated understanding, I would say it represents the spatial distance between two events which may or may not have happened simultaneously. Is this correct?

10. Feb 24, 2014

### ghwellsjr

That's correct. I think you would be better off using the standard non-delta version of the Lorentz Transformation and transform both ends of the length or time interval, then you can see if the coordinates in the new frame have the same Coordinate Time (for length) or the same Coordinate Distance (for time). It is most easiest to see the significant parameters if you plot them on a spacetime diagram. Have you ever done this?

Last edited: Feb 24, 2014
11. Feb 25, 2014

### scinoob

I haven't. I have something like a 5% understanding of Minkowski space, but I am in the process of studying it. I also have a somewhat good intuition about a light cone, but still far from what I want.

Thank you for the suggestion!

12. Feb 25, 2014

### ghwellsjr

Are you sure your understanding of Minkowski space is only at 5%? It's really very simple. It just means flat spacetime, as opposed to the curved spacetime of General Relativity. If you limit your concepts to three normal orthogonal spatial coordinates and add one for time, then that's all you need to start.

You should also limit yourself to scenarios that involve only motion along one spatial axis (usually considered to be the x-axis) and just draw one frame per diagram and use units where c=1, such as seconds and light seconds or nanoseconds and feet. I have already drawn a great many such diagrams for all kinds of such scenarios so if you want to get a jump start on understanding how to do it, just do a search on my name with "diagram" as the search term.

13. Feb 25, 2014

### scinoob

Yes, I do understand it that much. So I guess it could be more than 5% after all. I think what you're suggesting would be very helpful indeed, I like to visualize things to get a better intuition.

Thanks again!