The terms length contraction and time dilation

[Fredrik]

The terms "length contraction" and "time dilation"

Is there a particular reason why we say length contraction but time dilation? A Lorentz transformation \Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix} takes \begin{pmatrix}1\\ 0\end{pmatrix} to \gamma\begin{pmatrix}1\\ -v\end{pmatrix}, which dilates the time coordinate by a factor of \gamma, but the same \Lambda also takes \begin{pmatrix}0\\ 1\end{pmatrix} to \gamma\begin{pmatrix}-v\\ 1\end{pmatrix}, which dilates, not contracts, the position coordinate by a factor of \gamma. Of course a dilation by k is a contraction by 1/k, but the terminology still sounds weird to me.

(The upper component of the 2x1 matrices is the time coordinate).

 

[JesseM]

If you have two events at the same position but different times in one frame S, then in a different frame S' moving relative to S, the time between them will be greater, so that's time dilation. It is likewise true that if you have two events at the same time but different positions in one frame S, then in a different frame S' moving relative to S, the distance between them would be greater as well. But the "length" in length contraction does not refer to the distance between a single pair of events which may or may not be simultaneous depending on your choice of frame--rather it involves looking at two worldlines representing the front and back of an inertial object, with each frame defining the object's "length" by looking at the distance between a pair of events on the two worldlines that are simultaneous in that frame (i.e. simultaneous measurements of the distance between the front and back of the object).

 

[starthaus]

Is there a particular reason why we say length contraction but time dilation? A Lorentz transformation \Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix} takes \begin{pmatrix}1\\ 0\end{pmatrix} to \gamma\begin{pmatrix}1\\ -v\end{pmatrix}, which dilates the time coordinate by a factor of \gamma, but the same \Lambda also takes \begin{pmatrix}0\\ 1\end{pmatrix} to \gamma\begin{pmatrix}-v\\ 1\end{pmatrix}, which dilates, not contracts, the position coordinate by a factor of \gamma. Of course a dilation by k is a contraction by 1/k, but the terminology still sounds weird to me.

(The upper component of the 2x1 matrices is the time coordinate).

Yes, because the derivation takes us to:

dx'=dx/\gamma
dt'=dt\gamma

 

[cepheid]

An object is moving relative to me. The time interval I measure is greater than the elapsed proper time for that object (dilated).

The length I measure for that object is less than its proper length (contracted).

 

[JesseM]

Yes, because the derivation takes us to:

dx'=dx/\gamma
dt'=dt\gamma

You're missing the point of the original question I think--it really depends on how you define dx and dt. dt refers to the time in the unprimed frame between two events which occur at the same position but different times in the unprimed frame, with dt' as the time between the same events in the primed frame, so intuitively it might seem that the most analogous way to define dx in the length contraction equation would be as the distance in the unprimed frame between two events which occur at the same time but different positions in the unprimed frame, with dx' as the distance in the primed frame between the same two events. If you defined it this way, the correct equation would be dx' = dx\gamma. It's only because the "length" that appears in the length contraction equation is not defined in such an "analogous" (which is probably why the length contraction equation is usually written with symbols L and L' rather than dx and dx') that you end up dividing by gamma rather than multiplying by it.

 

[cepheid]

so intuitively it might seem that the most analogous way to define dx in the length contraction equation would be as the distance in the unprimed frame between two events which occur at the same time but different positions in the unprimed frame, with dx' as the distance in the primed frame between the same two events.

but instead dx is defined as the distance between two events which occur at the same time but at different positions in the primed frame, which is not an analogous definition. Is that what you're getting at?

 

[starthaus]

You're missing the point of the original question I think--it really depends on how you define dx and dt. dt refers to the time in the unprimed frame between two events which occur at the same position but different times in the unprimed frame, with dt' as the time between the same events in the primed frame, so intuitively it might seem that the most analogous way to define dx in the length contraction equation would be as the distance in the unprimed frame between two events which occur at the same time but different positions in the unprimed frame, with dx' as the distance in the primed frame between the same two events. If you defined it this way, the correct equation would be dx' = dx\gamma. It's only because the "length" that appears in the length contraction equation is not defined in such an "analogous" (which is probably why the length contraction equation is usually written with symbols L and L' rather than dx and dx') that you end up dividing by gamma rather than multiplying by it.

I don't think so, I wrote several posts on the same subject.

dx'=\gamma(dx-vdt)
dt'=\gamma(dt-vdx/c^2)

To measure dx' you need to mark the endpoints of the object in the comoving frame F' simultaneously, so, you need to make dt'=0.
This means dt=vdx/c^2. Substitute back in the expression for dx' and you get dx'=\frac{dx}{\gamma}

For time dilation, we want to get dt' as a function of dt when dx=0. You get immediately dt'=\gamma dt

 

[JesseM]

but instead dx is defined as the distance between two events which occur at the same time but at different positions in the primed frame, which is not an analogous definition. Is that what you're getting at?

Well, I was actually thinking that each frame would naturally tend to use their own distinct pair of simultaneous measurements of endpoints, so we wouldn't be talking about the distance between a single pair of events in different frames at all. But it is true that in the object's own rest frame, you can define its "length" with non-simultaneous measurements, so if you pick two events on the worldlines of either ends of the object which are simultaneous in the primed frame that sees the object in motion, the distance between these same two events in the unprimed frame where the object is at rest will qualify as the object's "length" in the unprimed frame even though they aren't simultaneous in this frame.

 

[Fredrik]

Thanks guys. I think Jesse "wins" this thread, because #2 got me to draw a spacetime diagram and see the difference between what I did in #1 and length contraction. The calculation I ended up doing looked like this:

L=x_A-vt_A=\gamma(vt_A'+x_A'-vt_A'-v^2x_A')=\gamma(1-v^2)x_A'=\frac{L_0}{\gamma}

(Sorry, I'm too lazy to draw the diagram that really explains what I'm doing. The point is that length contraction is more than just a Lorentz transformation).

I still think the terms "time dilation" and "length contraction" are pretty weird, because it sounds like length contraction would be to the x-axis what time dilation is to the t axis, when in fact length contraction has an "extra feature" that time dilation doesn't have. But it's not the only thing in physics that has a misleading name, and I think I can live with it.

 

[starthaus]

Thanks guys. I think Jesse "wins" this thread, because #2 got me to draw a spacetime diagram and see the difference between what I did in #1 and length contraction. The calculation I ended up doing looked like this:

L=x_A-vt_A=\gamma(vt_A'+x_A'-vt_A'-v^2x_A')=\gamma(1-v^2)x_A'=\frac{L_0}{\gamma}

(Sorry, I'm too lazy to draw the diagram that really explains what I'm doing. The point is that length contraction is more than just a Lorentz transformation).

I still think the terms "time dilation" and "length contraction" are pretty weird, because it sounds like length contraction would be to the x-axis what time dilation is to the t axis, when in fact length contraction has an "extra feature" that time dilation doesn't have. But it's not the only thing in physics that has a misleading name, and I think I can live with it.

Post 7 tells you in math what post 2 tells you in words.

 

[Rasalhague]

We had an epid thread about this last year! I think we more or less agreed that it can be a bit misleading when they're paired together in introductory texts as if contraction is the thing that space does, and dilation the analogous thing that time does.

 

[espen180]

Thanks guys. I think Jesse "wins" this thread, because #2 got me to draw a spacetime diagram and see the difference between what I did in #1 and length contraction. The calculation I ended up doing looked like this:

L=x_A-vt_A=\gamma(vt_A'+x_A'-vt_A'-v^2x_A')=\gamma(1-v^2)x_A'=\frac{L_0}{\gamma}

(Sorry, I'm too lazy to draw the diagram that really explains what I'm doing. The point is that length contraction is more than just a Lorentz transformation).

I still think the terms "time dilation" and "length contraction" are pretty weird, because it sounds like length contraction would be to the x-axis what time dilation is to the t axis, when in fact length contraction has an "extra feature" that time dilation doesn't have. But it's not the only thing in physics that has a misleading name, and I think I can live with it.

In your spacetime-diagram, if you construct a line parallel to the ct'-axis through x'=1, it will cross the x-axis at x=1/γ. In this sense we can have x'=γx. If you do teh same procedure on the ct and ct'-axes; construct a line parallel to the x'-axis through ct'=1, it will cross the ct-axis at ct=1/γ

In this way you can have the symmetry

x'=γx
t'=γt