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As an object moves faster, its time runs slower by factor X. I get why this is the case, but I'm not sure why distance is decreased by factor X as well.

Questions:

1) Why does a moving object's travel distance decrease by factor X as its time decreases by factor X? The explanation I've heard is that because of v = d/t, since the moving object's speed is constant from different reference frames, its distance must decrease as time decreases. But I thought motion was relative?

2) Why does the object's physical length, from the reference frame of an observer, decrease by factor X? Even if the object travels a smaller distance from part 1), I'm completely lost on how that would affect its length.

Also, I'm aware of the unification between space and time in General Relativity, but I'm learning about Special Relativity and would prefer an explanation in that realm.

Inertial Ignorance
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    I fly past you in my spaceship, which I consider to be 1 meter long. The front of my ship passes you at an event $A$ to which be both agree to assign time $0$ and location $0$. Consider an event $B$ which occurs at the back of my ship, simultaneous with $A$ according to me, and an event $C$ which occurs at the back of my ship, simultaneous with $A$ according to you. (It might be helpful to draw the worldlines of the front and back of my ship.) Compute the coordinates of $B$ and $C$ in our respective frames. What do you discover? – WillO Feb 15 '18 at 01:11
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    To be clear, time dilation does not cause length contraction. They are linked effects as a result of, as explained by @Chris, the nature of the invariant interval in space-time in SR - i.e. that that value involves both spatial dimensions and time in a particular way. – StephenG - Help Ukraine Feb 15 '18 at 01:46

3 Answers3

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A core tenet of special relativity is that the spacetime interval between two events:

$$ \Delta s^2=(c\Delta t)^2-\Delta\vec x^2 $$

is a Lorentz invariant (i.e. it's the same if measured in any two reference frames).

Since you know that $\Delta t^2$ varies between two frames due to time dilation, it follows that $\Delta\vec x^2$ must also vary in order to keep $\Delta s^2$ invariant. This is length contraction.

Chris
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  • No, I do not think that length contraction does in fact refer to different measurements of $\Delta x$^2. Let $A$ be the back of my spaceship at some moment. Let $B$ be the front of my spaceship at that same moment, according to me, and let $C$ be the front of my spaceship at that same moment, according to you. (Where you are in relative motion with respect to me.) Let $\Delta x_1$ be the distance I measure from $A$ to $B$ and $\Delta x_2$ be the distance you measure from $A$ to $C$. The ratio of these is the length contraction --- (CTD) – WillO Feb 15 '18 at 23:37
  • (CTD) whereas by contrast, you are looking at the interval between two fixed events (say $A$ and $B$), and observing that you and I will assign different values of $\Delta x$ to that pair of events. But length contraction is only partly that; it also incorporates the fact that we are choosing different pairs of events to measure between. (So that $\Delta x_1/\Delta x_2$ is in fact the wrong measure of length contraction.) – WillO Feb 15 '18 at 23:39
  • To put this another way: Distance contraction (between two fixed events) and length contraction are two different things. You are looking at the interval between two fixed events, and hence describing distance contraction, not length contraction. – WillO Feb 15 '18 at 23:41
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You say:

I get why time dilates

But in what sense, do you "get"? If you know that time dilation is a result the transformation laws between inertial frames being Lorentz transformations and not Galilean, then length contraction is another result from the same transformation equations. Time dilation doesn't cause length contraction, though they are related.

PhyEnthusiast
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  • I understood time dilation from a thought experiment I read. It involved a spaceship going past you at a high speed, and a light beam being shot from the ground of it to its ceiling, and then reflected back to the ground via a mirror. I would see the light beam moving in a triangle line, whereas the pilot on the ship would see it going up and down. Since the speed of light is constant, if the light beam travels less distance for the pilot, time must be running slower. However, would this also explain length contraction? Since the light beam travels less distance on the high speed ship. – Inertial Ignorance Feb 15 '18 at 09:01
  • I think you should add that in the question for clarity – PhyEnthusiast Feb 15 '18 at 10:35
  • @Inertial Ignorance no it does not. Because there is no contraction along the direction perpendicular to the direction of motion of the spaceship – PhyEnthusiast Feb 15 '18 at 10:41
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This is another "deep space" scenario but a little different. You're in a space station when a ship going 0.5c, or 150 million meters/sec, approaches. You know the ship has a length of 150 meters and it has a central master clock with large numeric displays on the sides, one at the nose and one at the tail. Because you have extremely sharp eyes you're able to read the clock's time on each display as the ship flashes by. You expect to read a time difference of exactly one microsecond, but, because the ship's clock is running slower than yours, you don't see 1 microsec but 0.866 microsecond as the time difference. And going 150 million m/sec, that makes the apparent length 130 meters instead of 150.

Pete Carter
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