# Are length and time Trade-offs in relativity?

1. Dec 9, 2015

### Dopplershift

In special relativity,
When an object travels at speeds near the speed of light, time slows down for the person on traveling at that speed, thus you gain more time. Also, you have length contraction.

So my question is this, with the idea of 4-dimesional space-time, is what you gain in time lost in length?

I know this is a strange question, but I'm trying to understand how the two are related.

2. Dec 9, 2015

### axmls

I'm not exactly sure what you mean by the phrasing of the question, but I'll answer with this:

When we measure the number of muons that are hitting the earth from cosmic rays bombarding the atmosphere, we find a lot more than we would expect to. They should have decayed by the time they reach the earth, but in reality, they're traveling so fast, that their clocks move slower, and their decay takes longer in our reference frame.

So, if an observer were watching a muon make it to the surface of the earth, they would say that the muon made it to the surface because its time was slower than ours, so it decayed slower.

If you look at the reference frame of the muon, however, this doesn't make sense, because it's at rest in its frame, so it should still decay at the same rate. However, the muon can easily say that it is a rest, and it is the earth that is traveling towards it. Then, the distance to the earth is contracted. So, in the muon's frame, it is able to make it to the earth because the distance was contracted, but time went at its normal rate.

These two effects perfectly cancel each other out so that both explanations are correct. The earth observer says the muon makes it to the earth because its clock was moving slower. The muon says that it makes it to the earth because the distance it had to travel was shorter.

3. Dec 9, 2015

### Dopplershift

@axmls,
That is a great explanation. So, let me see if I got this correct. In terms of length contraction, the muon sees the distance between Earth and itself shorter. But would the observer see the muon's length contracted, not its path, but the actual muon.
In other words, if I was on a space ship traveling at some speed, let's say 0.86c, I would see that the path would be shorter from point a to b, but wouldn't my spaceship be shorter to an outside observer?

4. Dec 9, 2015

### jbriggs444

This picture is not a good one, but a conclusion like the one that you are after can be derived.

When an object travels at near the speed of light, time does not slow down. Nor does length contract. Clocks on the object tick away merrily at one second per second -- as compared with other clocks on the object. Meter sticks on the object are still one meter long -- as compared with other meter sticks on the object. All the usual rules of physics apply in their usual way. There is no way for the object to tell whether it is really moving at nearly the speed of light or whether it is standing still.

It is when you compare clocks mounted on the moving object against [sets of synchronized] clocks at rest in some reference frame that you could conclude that clocks on the moving object are running slow. Is is when you compare [synchronized positions of the ends of] meter sticks on the moving object against a meter stick in some reference frame that you can conclude that meter sticks on the moving object are shortened.

How rapidly a clock ticks and how long a meter stick is are both relative. They depend on what frame of reference you use to measure them against. Hence the name "relativity".

However, not all things are relative. Some things are "invariant". An invariant quantity is one where you get the same result no matter what reference frame you use to perform your measurements. A place where you can find "trade-offs" is in the invariants.

One invariant is the length of a space-time interval. This is the separation between two "events". An event is a position in space at a particular time. It takes three coordinates of space (e.g. x, y and z) to specify the position and along with one coordinate of time (e.g. t). So you can specify an event with four coordinates. They won't be the same coordinates in every reference frame. But no matter what reference frame you use, there will be a set of coordinates for the event.

The length of a space time interval from (x1, y1, z1, t1) to (x2, y2, z2, t2) is computed as $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2 - (t_1-t_2)^2}$

Note the similarity the Pythagorean distance formula. This one is the same, except that there is a minus sign on the time coordinate.

It turns out that interval length is invariant. It does not matter what reference frame you use to measure the (x, y, z, t) coordinates for each event. The resulting computed interval length will be the same. If the distance offset between the endpoints increases and the time offset between the endpoints also increases, that is a trade-off that can make the length of the interval stay the same.

5. Dec 9, 2015

### Staff: Mentor

Yes, but the experiment is not sensitive to the muon's length in any way. Length contraction and time dilation are always present, but not always important.

6. Dec 10, 2015

### sweet springs

In rotation there is trade-off of longitudinal length and tansverselength. One is up and the other is down.
In Lorentz transformation trade-"on" of time and length takes place. When one is up, the other is also up.

7. Dec 10, 2015

### Mister T

When an object travels at near light speed relative to you, the traveler's clock runs slow by some factor relative to yours. The traveler's distance is reduced by that same factor, relative to you.

So, for example, at a speed of 0.866c the factor is about 2. You will observe the traveler's clock running at half the rate of your clock. The traveler will observe the distance traveled to be half what you observe. The math works like this. Suppose the proper distance traveled (length measured by you) is $L_o$ and the proper time (time measured by traveler) is $t_o$. Then the speed $$0.866c=\mathrm{\frac{L_o}{2t_o}}.$$
You can think of $\frac{L_o}{2}$ as the contracted distance for the traveler, or $2t_o$ as the dilated time for you, but not both because each effect happens in a different reference frame. So loosely speaking either you gain time or he loses length, but you don't combine those two effects as they are occurring to different people. No one of the two experiences them both.

This is really just another way of looking at the muon example.

8. Dec 10, 2015

### A.T.

Here is an interactive applet, which visualizes these relationships geometrically:

9. Dec 10, 2015

### axmls

I'll add the following to my post about muons: there are other reference frames where the muons make it to earth due to a combination of length contraction and time dilation, depending on how fast those frames are traveling. In other words, how much of the effect is due to length contraction and how much of it is due to time dilation depends on your frame of reference, and it's always possible to find frames where the effect is entirely due to time dilation, or frames where it is entirely due to length contraction, or frames where it's a little of both. In this way, you can essentially say that time dilation and length contraction are the same effect, but viewed from different reference frames.

10. Dec 11, 2015

### Jilang

Yes, from a given reference frame there is a trade-off. The further something travels through space in a given frame the less time elapses in that frame. The twins "paradox, is a good example of this. It's harder to see it with the muon example.

11. Dec 14, 2015

### Dopplershift

Thanks for all the insightful answers! :)