Confusion about relativity and length contraction and time dilation?

Question

So I know the equations of time dilation and length contraction given by: $$L' =\frac{1}{\gamma}L_0$$ $$t' = \gamma t_0$$

However, I am confused about how the speed of light stays the same in all reference frames through these equations. Imagine a rod moving relative to me at such a speed that $L' = 3 \cdot10^8$m and we know that the rod at rest has length $L_0 = 6\cdot10^8$m. Now if I see light go from one end of the rod to the other I record the time to be $1$ second. But In the reference frame of someone moving along at the same speed as the rod, they would see the light take $2$ seconds to go from one end to the other.

Therefore my time $t' = \frac{1}{2}t_0$ as something that took $1$ second to happen for me took $2$ seconds to happen for the other observer. But would this not make the equations $$L' = \frac{1}{2}L_0 \implies L' = \frac{1}{\gamma}L_0$$ $$t' = \frac{1}{2}t_0 \implies t' = \frac{1}{\gamma}t_0$$

I don't get what I did wrong to get the wrong equations.

Answer 1

When you will-nilly apply time-dilation and length-contraction powers of $\gamma$:

Confusion will result.

I know this because I lived in this confusion for many years.

You really need to talk about events:

$$ E = (t, x), $$

which are frame-less points in space-time, that's key: they have no frame, they don't move. In all frames, they are a blink at $x=E_c$ when the clock says $t=E_t$, or at $x'=E'_c$ when $t'=E'_0$.

So, let's look at a rod of length $L$ at rest, in frame $S$. Its endpoints follow the world lines:

$$ W^- = (t, 0) $$ $$ W^+ = (t, L) $$

If we measure then length at $t=t_0$, then:

$$ L = (W^+_x - W^-_x) = L - 0 = L $$

That's pretty easy. Let's say $t_0=0$, then the measurement events are:

$$ E^- = (0, 0) $$ $$ E^+ = (0, L) $$

which makes sense. Now we can transform those event to a frame $S'$ moving at $\beta = v/c$ (we'll do $c=1$):

$$ E'^- = \big(\gamma(t^--vx^-), \gamma(x^--vt^-) \big)' = (0, 0)$$

$$ E'^+ = \big( -vL, \gamma L \big)' $$

So as far as $S'$ is concerned, $S$ did not measure the endpoints at the same time. He measured the leading endpoint later, so ofc. he got a longer answer than we, the moving frame, did.

(Note: the primes on the event don't mean anything. An event is an event, I just put a prime on it to remind you that I am representing it in the coordinates of the primed frame: $E = E' = E''$, etc, always, as events are fixed points in space-time).

Now you can solve this problem algebraically, but since I know the answer. I'll just state it. I need to introduce a third event. We need to measure the ends at:

$$ E'^- = (0, 0)' $$ $$ E'^{+'} = (0, L/\gamma)' $$

for a length of:

$$ L' = E'^{+'}_x - E'^-_x = L/\gamma - 0 = L/\gamma $$

Length Contraction.

If I transform those back to $S$:

$$ E^- = (0, 0) $$ $$ E^{+'} \big( -vL', L \big) $$

so we measured the leading edge before the training edge, so ofc it is shorter.

tl;dr: when in doubt, do the Lorentz Transform.

Answer 2

You are making at least two basic mistakes.

Firstly, you are misusing the time dilation formula. That formula only applies where you have two events that occur in the same place in one frame, which is not true in your example.

Secondly, you are overlooking the fact that the ruler is moving in your frame. The length contraction formula tells you how far apart are the two ends of the ruler at a given time in your frame. In the example you site, the light takes time to travel from one end of the ruler to the other, so you are not considering the positions of the two ends at the same time. For instance, suppose you label the points A and B which are the positions of the two ends of the ruler at some point in time in your frame, and that they are one light second apart. Light will indeed take one light second to go from A to B, but by the time it reaches B, the end of the ruler will be somewhere else!

Answer 3

The length contraction formula describes how two frames disagree on the length of an object, and the object must be at rest in one frame. Light is not at rest in any frame, so the length contraction formula doesn’t apply.

The time dilation formula describes the ratio of the proper time to coordinate time for a clock. There is no clock that moves at the speed of light, so that formula also doesn’t apply.

The formula to use is the Lorentz transform. That formula will always work for inertial frames in flat spacetime. That should be what students start with. They will automatically simplify to the time dilation and length contraction formulas when appropriate.

Answer 4

Moving frames in special relativity are not Galilean:

I think a lot of confusion can be generated if the popular narratives of special relativity are tried to be interpreted on a Galilean framework.

It is often said, that according to special relativity, time runs slower in a moving reference frame. This sound a bit like a Galilean reference frame with rescaled time. I.e., $$ (t, \vec x) \mapsto (t', \vec x'):= (\alpha t, \vec x + t \vec v)\,,$$ where $\alpha\in (0,1)$ is some scaling factor. Sometimes it is also said, that length is contracted, which sound like $$ (t, \vec x) \mapsto (t', \vec x'):= (\alpha t, A(\vec x + t \vec v))\,,$$ where $A$ is some scaling matrix, e.g., $A = \begin{pmatrix}\lambda_1 & 0& 0 \\ 0 &\lambda_2 & 0 \\ 0&0& \lambda_3\end{pmatrix}$, with $\lambda_i \in (0,1)$.

However, this is not how special relativity works! It is easy to see, that rescaling time and space independently cannot explain the phenomena of special relativity like constant speed of light. E.g, consider the speed of light in a moving train and light ray coming from the front and one coming from behind. Rescaling $t'$ could not make the speed of both light rays independent of the train's velocity $\vec v$. Many other thought experiments like this may be cooked up.

How special relativity works:

The crucial difference in special relativity is that space and time coordinates are mixed, while $t'$ in non-relativistic physics only depends on $t$, in special relativity it depends on both $t$ and $\vec x$. This has the consequence that the notion of simultaneity of different reference frames do not agree.

At this point, it is more helpful to switch to a spacetime picture instead of constantly switching reference frames. If you draw some space time diagrams it is usually rather clear what is going on.