# If time runs slow for moving train, how can it cover same distance in less time than someone on the ground unless the track decreases in its view?

According to special relativity,

Δt=γ*Δt' ... (1)

Where,

Δt is the time the train takes to completely pass by me according to my watch,

Δt' is the time I should see has passed for someone on the train.

If v is the speed of the train, multiplying both sides of eqn (1) by it,

L=γ*L' ... (2)

Here,

L= length of train as I would measure it,

L'= length of train for someone on the train (¿ but from my point of view).

Or,

L'=L/γ,

To my understanding, what the equation as I have written it is saying is that, from my point of view, someone on the train should "measure" lengths to be smaller than I would measure it. It also explains how in "less" time than in my watch, the train could cover the distance of the platform; that is if the platform's length decreased for the train.

But according to every other sources a moving train will appear to be smaller in length for someone on the ground than what someone on the train measures. But equation (2) tells the opposite of it.

Where am I going wrong? Please someone explain this for me.

Your question is essentially a perfect example of why I recommend against using the length contraction or time dilation formulas at all. Instead, students should be taught to use the Lorentz transform. The Lorentz transforms automatically simplify to the length contraction formula when appropriate, and in the appropriate form. So here to use the Lorentz transform I would write the worldline of the front and back of the train as $$F=(ct,vt)$$$$B=(ct,vt-L)$$We then apply the Lorentz transform to obtain $$F'=\left(\frac{ct}{\gamma},0\right)$$$$B'=\left(\frac{ct}{\gamma}+\frac{Lv\gamma}{c},-L\gamma\right)$$ So in the primed frame note that the length of the train is $$L'=L\gamma$$, which is backwards from what you had thought. Note also that the time for the clock in the front of the train is $$ct'=ct/\gamma$$. Note finally that due to the relativity of simultaneity, the time in the rear of the train is offset by $$Lv\gamma/c^2$$.

It requires all three effects, length contraction, time dilation, and relativity of simultaneity to explain this scenario. And it is important that each effect be applied the correct way. Again, I recommend against using the time dilation and length contraction formulas. Just use the Lorentz transform directly.

As Dale says, the time dilation formula is a special case and is only applicable in a particular type of situation, so where you have gone wrong is in mis-applying it at Equation 1).

The time dilation formula does not mean that a single moving clock ticks at a slower rate than a single stationary one. It applies only where you are comparing readings on a single clock in one frame with two separate clocks it passes in another. What it says is that the elapsed time shown on the single clock is always less than the difference in the readings between the two separate clocks it passes. Let me repeat that- the formula applies where you have one clock passing two separate clocks, and the interval measured by the one clock is always lower.

If you apply that to the situation in which you are sat on a platform and you measure the time it takes for a train to pass you, then you are the one clock. On the train you would need two clocks- one at the front and one at the rear- to measure the time it takes you to pass the train. So according to the time dilation formula, it is you who are time dilated!

By applying the formula the wrong way around at the start, all of your subsequent steps in reasoning- which I think are fine, by the way- just propagate the mistake and take you to the wrong conclusion.

I would go further than Dale, and ban the use of the phrase 'moving clocks run slow', as I think it is responsible for half the misunderstandings in special relativity!

The key for understanding is to separate coordinate time from proper time. For Alice in the train looking at her clock, it is proper time . If we place several synchronized clocks at the platform it is coordinate time (in the platform frame). Now, if Alice compares her clock with each platform clock when they passes in front of her, she notes that the time intervals in the platform are longer than at her clock.

That is the meaning of $$\Delta t = \gamma \Delta t'$$

As she sees the platform travelling with velocity $$v$$, the travelled distance between two clocks in the platform is $$vt'$$. She measures any distance there shorter.

But the same reasoning could be made for Bob standing at the platform. Now each wagon of the train has a synchronized clock in the train frame. That is the coordinate time (for the train frame). Bob, looking always to his own clock, measures proper time. When he compares his clock with the wagon clocks when they passes by him, he notes that the time interval of that wagon clocks is greater than his own.

When he multiplies the time interval (from the train front until the train tail passes by him) with his clock by the same velocity $$v$$, the train length will be shorter than for Alice in the train frame.

Just to be clear, you highlighted your exact problem when you wrote the question in “length of train for someone on the train (¿ but from my point of view).” Τhere is a mixed perspective happening here.

You have correctly defined a $$\Delta t$$ as the time between two objective events, one of them being the front of the train passing you, and one of them being the back of the train passing you. You have correctly calculated that during this time interval if you are watching someone in the train holding a clock, their clock only increments by $$\Delta t/\gamma$$ between these two events. And you have correctly calculated that you think the length of the train is $$v\Delta t$$ but now you incorrectly transfer to the train’s perspective.

Why is this incorrect? Because of the relativity of simultaneity.

Put a clock at the front and back of the train, that the person in the train thinks are synchronized. You, on the ground, do not think they are synchronized! Rather you see them both tick at $$1/\gamma$$ seconds per second and also the one at the front of the train always seems to be earlier than the one at the back of the train by a fixed offset $$\sigma$$. And here we have to be a little bit clear, $$\sigma$$ is just the difference between the two passing numbers on the clocks, at a time that I on-the-ground think is simulataneous.

To understand what the person in the train thinks happened, we can then just write down the times of these two clocks as they pass you. So the one at the front initially shows $$t'$$, then the one at the back shows $$t' + (\Delta t/\gamma) + \sigma,$$ and so according to the person in the train, the time it took you to pass by their entire train was $$(\Delta t/\gamma) + \sigma,$$ because they think these clocks are synchronized... and their calculation of the length of the train is therefore longer than you thought, by a distance $$v~\sigma$$.

What is this offset $$\sigma$$ more concretely? It is $$L_0 v/c^2$$ where $$L_0$$ is the length that the person on the train thinks the train is.[1]

If you are thinking like an ordinary person you will now say, “But, that's what I wanted to calculate! You can't tell me that to derive $$L_0$$ I first have to know $$L_0$$!”. But we think like physicists and do this all the time, just set up the equation: \begin{align} L_0 &= v{\Delta t\over\gamma} + v\sigma\text{, where } \sigma = \frac{vL_0}{c^2},\\ L_0 &= v{\Delta t\over\gamma} +L_0\frac{v^2}{c^2},\\ &\text{(collect } L_0\text{ terms on the left})\\ \frac{L_0}{\gamma^2} &=v{\Delta t\over\gamma}\\ L_0&=\gamma v \Delta t,\end{align}

which was to be proven.

The technical term for thinking this way is to say that there is only one “self-consistent solution” or so.

Note 1: This math can get a bit subtle. Don't get caught up putting a $$\gamma$$ in this expression $$L_0v/c^2$$ by using the Lorentz transform directly, because the land-time that you transform the train-time zero to will not be $$t=0$$ and that messes everything up. Instead you want to Lorentz-boost the land-coordinates $$(t, x) = (t, vt-L)$$ into the train frame to get $$(t/\gamma +\gamma L v/c^2, \gamma L)$$, versus boosting just $$(t, vt)$$ at the front of the train to get $$(t/\gamma, 0)$$. The distance between these lines at constant time is now super easy: the rest length is the de-contracted $$L_0=\gamma L$$, while at $$t=0$$ the clock at the back of the train must have been showing $$t'=+L_0 v/c^2,$$ which is our $$\sigma.$$ (Of course if you do it this way then you already prove the length contraction formula en route to providing the fact $$\sigma=L_0v/c^2$$ that explains the answer... c'est la vie.)

Special relativity is counter intuitive. The basic reason is that time is different than classical physics leads you to think. It takes a long time to get used to this.

To measure the length of a train, you have to measure the position of the head and tail at the same time. In classical physics, this is not a problem. Everybody agrees on what time it is. You just need to measure the positions.

In special relativity two people may agree on the time when the tail of a train passes you, and disagree about the position of the head at that instant.

Suppose a platform in a train station is just as long as the train when the train is at rest. Suppose the train is moving. You are standing at the beginning of the platform. In your frame, the train is foreshortened. The head of the train passes a flag partway down the platform at the same time as the tail passes you at the beginning of the platform.

The conductor stands in the caboose. As he passes you, he agrees on the position of the tail of the train. But at that instant, he says the head of the train has already passed the flag and the end of the platform. This is not appearance. The two passings of tail and head really do occur at different times in the conductor's frame, and at the same time in yours.

We are used to this idea for positions. I watch the conductor go by. At two different times I see him in two different positions. But the conductor says the train is motionless and the world goes by. He says he is in the same position at those two times. But applying this idea to time is mind bending.

One of the problems with learning special relativity is that it is taught in a way that most quickly gets to the correct math. Along the way, it turns out that the world is very different than you have always seen it to be. You are expected to just accept that the speed of light is constant, and all of these weird consequences are true. This approach can be justified because these weird consequences really are true. And they are hard to get used to regardless of how they are introduced.

Different ideas of simultaneity are behind the different lengths of the train. In the conductor's frame, he measures the position of the head and tail at the same time and comes up with the usual length of the train. The conductor says the head of the train passes the flag before the tail enters the station.

In your frame, time is different. The head passes the flag at the same time as the tail enters the station.

The conductor is not surprised you get a shorter length for the train. In his frame, you measured the position of the head before the tail arrived at the beginning.

The ladder paradox and the twin paradox are explained by similar disagreements. The lesson from them is often how weird special relativity is. They should be taken as examples of how the universe really behaves, and how you need to learn to think.

It might help if you knew why the speed of light is constant. Here are some answers. It may not help as much as you would like. Do we know why there is a speed limit in our universe?

I wan't to point out a fondamental but often forgotten principle: All physics should be captured by sticking with only one reference frame. You are free and recommanded to change to a reference frame where the problem looks simpler, but no jumping back-and-forth.

Always make sure the physical quantities you're using in dynamical equations are defined (measured) in the same frame at the same time.

Equations like Lorentz transformation which connect quantities according to different frames should be only used at the beginning or ending of calculations.

So, in the ground frame, the train has a velocity and a length measured by difference of coordinates of front and back at the same time. We also have the length of a part of track that you want the train to cover. The reading of a clock on the train is also an observable in the ground frame by comparing it to any nearby rest clock. However, 'time of the train frame' is not a reasonable observable, instead we have a set of readings for clocks placed at different positions of the train -- and we'll find they are not equal to each other, relativity of simultaneity.

Every detail you may concern could be deduced solely from analysis in the ground frame. You may wonder how to deduce reading of clocks on the train, the answer is dynamics -- reading is actually clock hand's position! The possibly simplest model for clock are a photon being reflected endlessly by two parallel mirrors. It's easy and fun to derive time dilation solely from dynamics in the ground frame. For more realistic and complex clocks, dynamical analysis is viable in principle and is guaranteed by the relativity principle. (Otherwise we could deduce from difference between different kind of clocks a superior reference frame. )

DO NOT get me wrong, Lorentz transformation has another extremely important usage: to check if a physical law is Lorentz invariant -- if it is a real physical law. Or to use it reversely, formulate invariant candidate for physical law. But when you start to use these laws, remember the principle!