# Time Dilation Doubt

1. Jul 20, 2015

### Raman Choudhary

Suppose there is person A in a frame S with respect to which a light beam clock is at rest , now time taken by a photon to go from bottom to top is L/c according to A.
now the same clock is looked at by a person B travelling with velocity v w.r.t S and according to him time taken by photon to reach from bottom to top is L/c*gamma.
In this proof how can we know that when A's clock showed L/c second at that very instant B's clock didn't show L/c*gamma becoz that would mean something opposite to time dilation.
I mean showing the times the photon takes w.r.t different frames barely gives me an idea about time dilation PARAPHRASE=how does this proof incorporate the fact that B's clock is slow????HOW DOES IT PROVE TIME DILATION

2. Jul 20, 2015

### cpsinkule

The clock in A's frame does not tell you what the actual time is (although all you need to do is synch a real clock up with it). In the derivation, frame B DOES NOT refer to any time calculations in frame A. All we are doing is calculating what each observer sees INDEPENDENTLY and then we compare them.
The two postulates of special relativity are covariance (inertial frames have the same physics) and universality of c regardless of relative motion. With these postulates it can be shown that time is dilated between inertial observers. In a rest frame, the photon bounces up and down in a straight line, 2L/c=τ being the period if the mirror separation is L. In a frame moving relative to this frame, the photon will appear to take a "zig zag" path. If Δt is the period of the clock in the frame moving relative to the stationary clock, then in time Δt the clock would have moved a horizontal distance of vΔt (v is the relative velocity). Now, vΔt is the base of the triangle formed from the full cycle of the photon bouncing. the two other sides are just the distance the light has traveled, and from the universality of the speed of light, this distance must be cΔt/2 for each side. Of course, the height of this triangle is just the separation of the mirrors, L. All you need to do now is apply basic geometry of right triangles. a2+b2=c2

Drop a perpendicular from the vertex of the triangle to the base. Two right triangles will be formed. The base is just vΔt/2, height is L, and hypotenuse is just cΔt/2 so

(vΔt/2)2+l2=(cΔt/2)2
or
(vΔt/2)2-(cΔt/2)2=L2
factoring out (cΔt/2)2 from both terms on the left side of this equation and taking the square root, we find
cΔt/2√(1-v2/c2)=L or Δt*1/γ=2L/c but 2L/c is just τ

Notice that, in deriving the time it takes for the photon to complete its motion in the moving frame, we made absolutely NO assumptions as to what the rest frame sees.

3. Jul 20, 2015

### Raman Choudhary

suppose γ=3 and t=2 where t is the time shown by A's clock when the photon reaches the bottom again.
now as we look at them individually what i interpret from the proof is that "both the clocks (i.e. A's and B's clocks) are ticking at the same rate i.e when A's clock reads 2 sec B's clock reads 2 sec. too but it's just that for B the photon will hit the bottom back after 4 more seconds(i.e total 6 seconds) and at that time A's clock will also show 6 sec. but it's just that the photon according to A had already hit the bottom at t=2sec."where in ur proof is it shown that B's clock runs slow.
that's my interpretation from the proof please tell me how this proof i.e. what u have written shows that B's clock runs slow??
prove my interpretation wrong.???

4. Jul 20, 2015

### cpsinkule

B and A see each other's clocks run slow.

5. Jul 20, 2015

### Raman Choudhary

HOW????

6. Jul 20, 2015

### Staff: Mentor

7. Jul 20, 2015

### Raman Choudhary

Sir i have doubt in this particular case(light beam clock) so the link u have given is irrelevant to my needs.

8. Jul 20, 2015

### Staff: Mentor

Raman Choudhary, have you looked in any relativity textbooks or other mainstream sources for an explanation of the light clock and how it works? Most of them discuss it. Asking "how does a light clock show time dilation?" is too broad a question; if you can find a specific presentation with a specific point you're not sure about, and ask about that, you're much more likely to get a useful answer.

9. Jul 20, 2015

### Raman Choudhary

Sir my doubt is the following,
suppose γ=3 and t=2 where t is the time shown by A's clock when the photon reaches the bottom again.
now as we look at them individually what i interpret from the proof is that "both the clocks (i.e. A's and B's clocks) are ticking at the same rate i.e when A's clock reads 2 sec B's clock reads 2 sec. too but it's just that for B the photon will hit the bottom back after 4 more seconds(i.e total 6 seconds) and at that time A's clock will also show 6 sec. but it's just that the photon according to A had already hit the bottom at t=2sec."where in ur proof is it shown that B's clock runs slow.
that's my interpretation from the proof please tell me how this proof i.e. what u have written shows that B's clock runs slow??
prove my interpretation wrong.???

10. Jul 20, 2015

### Stephanus

Perhaps it's better to show in a picture that I make before.
Don't worry if you don't grasp SR. I'm struggling, too

Supposed, a train, it's wide (not length!) is 0.5c
From your frame, the train is at rest.
If the train is not at the same frame with you who sit on a certain row, so the chair, desk, windows, everything will hit you inside.
And if you sit at F and light a signal to D and back to F, how many seconds pass according to you?
The length is 0.5c+05.c = c.
So, your watch shows 1 seconds, right?

11. Jul 20, 2015

### Staff: Mentor

Meaning, we are using an inertial frame in which the light clock is moving at a velocity such that $\gamma = 3$, correct? This means that $v = \sqrt{1 - \left( 1 / \gamma^2 \right)} = 2 \sqrt{2} / 3$.

Note also that you haven't specified in which direction the light clock is moving. I'll assume that it's moving in a direction perpendicular to the direction the photon inside the clock moves since that appears to be the case you're asking about. (The case where the light clock is moving parallel to the direction the photon inside the clock moves requires a different analysis.)

So A is the observer who is moving with the light clock (relative to him, the light clock is at rest), and B is the observer who is not moving with the light clock (relative to him, the light clock is moving), correct? So 1 unit of $t$ in A's rest frame is the time it takes for the photon to go from one mirror to the other inside the light clock.

What proof? Do you have a reference? You haven't given any proof (you give some vague statements that might be part of one, but they're too confusing for me to make sense of), but you seem to have read one. Which one?

Rather than try to guess what the proof you read said, I'll outline how I would analyze the case described above. In analyzing any scenario in relativity, the first step should always be to pick out the events of interest and assign them coordinates in some chosen inertial frame. Then, to find out how those events look in some other frame, we just use the Lorentz transformation to convert the coordinates from one frame to the other.

In this case, we'll start with A's rest frame. In this frame, we have three events whose coordinates are easily given:

Event #1: Photon leaves bottom mirror. Coordinates: $t = 0$, $x = 0$, $y = 0$. (We don't need the $z$ coordinate since only two spatial dimensions are involved.)

Event #2: Photon hits top mirror. Coordinates: $t = 1$, $x = 0$, $y = 1$. (Here we have chosen the $y$ direction as the direction the photon moves in A's rest frame. We use units in which $c = 1$, so the photon moves one unit of distance in one unit of time.)

Event #3: Photon returns to bottom mirror. Coordinates: $t = 2$, $x = 0$, $y = 0$.

Now all we need to do is transform these coordinates to B's rest frame. We use the Lorentz transformation formula, which says $t' = \gamma \left( t - v x \right)$, $x' = \gamma \left( x - v t \right)$, $y' = y$. So we get (using $\gamma = 3$ and $v = - 2 \sqrt{2} / 3$, from above--note the minus sign on $v$, we are assuming that the light clock moves in the positive $x$ direction relative to B, but here we need B's motion relative to the light clock, which will be in the negative $x$ direction):

Event #1: $t' = 0$, $x' = 0$, $y' = 0$. (Note that this event has the same coordinates in both frames. We call this event the "origin" of the frames.)

Event #2: $t' = 3$, $x' = 2 \sqrt{2}$, $y' = 1$.

Event #3: $t' = 6$, $x' = 4 \sqrt{2}$, $y' = 0$.

So according to B, it takes 6 units of time for the photon to make one round trip, instead of 2--a time dilation factor of 3, as expected. Since the speed of light is the same in both frames, the photon must also travel a longer distance, and indeed it does: if you compute the total distance the photon travels, you will see that it travels 6 units of distance. Everything fits.

12. Jul 20, 2015

### Stephanus

Sorry, should have brought a picture that conforms you calculation. But this is the only picture in my harddrive. Should redraw it and upload it again...

13. Jul 20, 2015

### Stephanus

I will answer what I know is true.
Perhaps the good mentors/advisors should step in if there's a mistake in my answer, so not to confuse Raman Choudhary
So, this is the more appropriate picture.

The wide of the train must be 300 thousands km. That's what time it takes for light 2 seconds to bounce from F to D and back to F.
For convenience, we say the length of Q is c in the unit of light second.
The speed?
Yep, $v = \frac{2\sqrt{2}}{3}$
What does it tell you from this picture?
If you are at the train, you'll see the light travels something like this:

The light from the train frame travels at 2c bounce back. So your clock must have shown 2 seconds if you are in that train.
But if you are at the platform, you'll see that the train travels at $\frac{2\sqrt2{}}{3}$, so you'll see the light forms two hypotenuses of two triangles

If you said $\gamma = 3$, (and PeterDonis has calculated that $v = \frac{2\sqrt{2}}{3}$),
So, $t = 2 * \gamma = 6$ seconds; $AF = vt = \frac{2\sqrt{2}}{3}3 = 2\sqrt{2}$
The length of p? Simply $0.5t = 0.5 * 6 = 3c$
Or using phytagoras.
$p^2 = c^2 + (2\sqrt{2})^2 = 1 + 8; p^2 = 9; p = 3$
So, from the platform frame, the light travels at 6c. So it takes 6 seconds for the train to travel from A to B according to platform frame.
This is the proof that clock at train frame runs slower than on the platform frame.
The proof that clock on the platform runs slower than the train??
I know, time dilation works both ways.
I'll have to figure it out, wait...

14. Jul 20, 2015

### Staff: Mentor

No, it doesn't, it travels at $c$. But it has to travel 6 light seconds in the platform frame, as opposed to only 2 light seconds in the train frame. So it takes 6 seconds to travel in the platform frame, vs. only 2 seconds in the train frame.

That will be a different clock, one at rest relative to the platform instead of the train.

15. Jul 20, 2015

### Stephanus

I'm sorry, I'm sorry. It's not a calculation mistake. It is a grossly typo, I assure you. Light travels at c(km/s) for 6 seconds for 6c(km) distance. That's what I wanted to type. But I typed quickly. Can't clearly state my mind.

Last edited: Jul 20, 2015
16. Jul 20, 2015

### Stephanus

Okay,..
I try to avoid ST diagram as possible,
This is the original light path as seen from the rest platform.

What does this tell us?

Train (T) will see that the light travels 2c distance. So when the train (T) arrives at B, T will see that T's clock is 2 seconds.
T will see that B's clock is 6 seconds, but don't confuse 6 seconds! Because the clock might no by synchronized at first.
But there is one thing that we should take for granted to understand this time dilation.
A's clock and B's clock must be synchronized!.
So when T arrives at B, they must confer, all of them A,B and T.
But A shouldn't come to B, even if A walks to B at 4mph, A's clock won't be synchronized with B even with negligible $\frac{1}{\sqrt{1-4mph/c^2}}$. Phone will be safe

A: "T starts travel at A:0 seconds and T's clock read 0"
B: "T arrives at B: 6 seconds and T's clock read 2"
So 2 seconds for the train is 6 seconds for the platform, that we already knew.

But what about the T? How can T prove that time dilation is mutual?
T should also have friends, too to confer with T, because train's clock and platform clock are differents and they might not be synchronized.
They are the cars.
T's clocks between cars must be synchronized!.
See this picture. What does it tell us?

Now, A also flashes it light along the rail width, perpendicular with the direction of the train, so we'll have the same situation.
T will see that A's light is slanted, of course T always see that T's light is straight, perpendicular to T's travel direction.
Now, when T Front (TF) arrives at B, TF will read its clock: 2 seconds.
How does TF see the light that's flashed by A? The light will be slanted, and 2 seconds is not enough for the light to bounce back from the other side of the rail to A.
In 2 seconds, the light only travels at the green part of the arrow, see picture.
T calculates the distance by pythagoras theorem.
The hypotenuse is, of course, 2 ls. Remember TF clock when it arrives at B read 2 seconds.
T speed is $\frac{2\sqrt{2}}{3} = 0.9428c; \gamma = \frac{1}{\sqrt{1-\frac{(2\sqrt{2})^2}{3^2}}} = 3$ from now on, we'll use decimals.
So in 2 seconds, A will be at 1.8856c.
Interestingly when TF arrives at B, A haven't reached Train Back/Stern (TB) car, somehow AB distance is length contracted.

After all is settled, the train confers.
So, TF consult one of his car in the middle that is 1.8856 ls away from TF, let's call this middle car TM.
Remember, the hypotenuse is 2 ls, horizontal distance is 1.8856 ls, so vertical distance (D) is $D = \sqrt{2^2-1.8856^2} = 0.667$ seconds. A will see that the light only travels 0.667 ls after TF left, or after A flashed the light, so A's clock is 0.667 seconds
TM: "A's clock is 0.667 seconds" This is the most important clock time in our discussion

----------------------------------------------------------------------------------------------------------------
TF thought, "Funny, I left when A's clock reads 0 seconds, now when my clock (TF) reads 2 second, A's clock reads 0.667 seconds. Subtracting 0.667 - 0 = 0.667 (always remember platform's clock might not be synchronized with trains clock), so platform's clock is dilated by a factor of 3"
B thought: "Interesting, when TF arrives, my clock (B) reads 6 seconds and TF reads 2 second, A reports that when TF left, TF's clock read 0 seconds. Subtracting 2-0 = 2 (clocks might not be synchronized), so train's clock is dilated by a factor of 3"
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This is the S-T diagram.
This diagram is not to scale!
V is 0.6 instead of 0.9428c. Because when you tranform it to rest, it disappear from the screen. Gamma factor is to big.
But the concept is the same.

As predicted, there's simultaneity of events.
At platform rest frame, E1 and E3 is at the same time.
At train rest frame, It's E1 and E6 is at the same time.
The maroon rectangle (TM) is the location of the middle car that is interogated by TF.

I hope I don't make any mistake and I hope the good mentors and advisors would be so kind to correct me immediately before the OP is mislead.
Thanks.

Last edited: Jul 20, 2015
17. Jul 21, 2015

### Raman Choudhary

Thank you everyone for your kind replies.
I still haven't got it , my bad luck.

18. Jul 21, 2015

### Stephanus

That makes the two of us . It takes me a long long time to understand it. But now, I can grasph, at least space time diagram. Altough only in x direction.
Let me try to guide you.
Do you know the two laws in relativity?

19. Jul 21, 2015

### Raman Choudhary

yes sir

20. Jul 21, 2015

### Stephanus

Two laws? What are those?