How does this proof demonstrate the concept of time dilation?

[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 traveling 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

 

[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.

 

[Raman Choudhary]

thank u for ur reply,

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.

thanks for ur reply,
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.?

 

[cpsinkule]

thank u for ur reply,

thanks for ur reply,
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.?

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

 

[Raman Choudhary]

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

HOW?

 

[Nugatory]

HOW?

You'll find many threads here explaining how A and B can both find the other clock to be slower than their own without contradiction. For example: https://www.physicsforums.com/threa...uns-slower-in-relativity.822850/#post-5165653

 

[Raman Choudhary]

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

 

[PeterDonis]

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.

 

[Raman Choudhary]

Sir my doubt is the following,
thanks for ur reply,
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.?

 

[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?

 

[PeterDonis]

suppose γ=3

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.)

and t=2 where t is the time shown by A's clock when the photon reaches the bottom again.

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 i interpret from the proof

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.

 

[Stephanus]

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 $v = \sqrt{1 - \left( 1 / \gamma^2 \right)} = 2 \sqrt{2} / 3$.[..]

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...

 

[Stephanus]

Can I answer?
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.

[..]
suppose γ=3 and t=2 where t is the time shown by A's clock when the photon reaches the bottom again.[..]

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?

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 $v = \sqrt{1 - \left( 1 / \gamma^2 \right)} = 2 \sqrt{2} / 3$.[..]

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...

 

[PeterDonis]

from the platform frame, the light travels at 6c

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.

The proof that clock on the platform runs slower than the train??

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

 

[Stephanus]

No, it doesn't, it travels at $c$. But it has to travel 6 light seconds in the platform frame, [..]

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.

 

[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.
TF: "TM!, please read A's clock"
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"
----------------------------------------------------------------------------------------------------------------------

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.

 

[Raman Choudhary]

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

 

[Stephanus]

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

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?

 

[Raman Choudhary]

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?

yes sir

 

[Stephanus]

yes sir

Two laws? What are those?

 

[PeterDonis]

I still haven't got it

Is there something specific that you can't follow?

 

[Raman Choudhary]

The two postulates of special relativity are covariance (inertial frames have the same physics) and universality of c regardless of relative motion.

Two laws? What are those?

 

[Raman Choudhary]

Is there something specific that you can't follow?

Sir i know that in this example of light beam clock we have to first calculate the times observed by different observors INDEPENDENTLY and then look for deductions.
Now after calculating the respective times i.e 2L/c and 2L/c*gamma INDEPENDENTLY; How shall i deduce from these that time has been dilated for one observor i.e w.r.t whom light beam clock moves horizontally.
As regards to what you told me about it being explained in any book, I have 2 books(Concepts of Physics by H.C VERMA and Fundamentals of Physics by HALLIDAY/RESNICK/WALKER.)in both of them they have calculated the times as observed in different frames and wrote in the next line that " the proof(which same as described in one of the replies above) clearly shows that the observor w.r.t whom the clock is moving has his clock ticking slower than the other observor".Maybe they both have their clocks ticking at the same rate but i know that is not true but why??
I think now my doubt is clear to you sir?

 

[PeterDonis]

after calculating the respective times i.e 2L/c and 2L/c*gamma INDEPENDENTLY; How shall i deduce from these that time has been dilated for one observor i.e w.r.t whom light beam clock moves horizontally

You've got this garbled. The light clock is moving with respect to observer B; that means that, to observer B, the light clock appears time dilated. It does not mean that observer B appears time dilated with respect to himself.

" the proof(which same as described in one of the replies above) clearly shows that the observor w.r.t whom the clock is moving has his clock ticking slower than the other observor".

This looks garbled too. The light clock is ticking slower than observer B's clock (relative to observer B), because there are six ticks of observer B's clock for every two ticks of the light clock.

 

[Janus]

Sir my doubt is the following,
thanks for ur reply,
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.?

Maybe the above animation will help. The two light clocks start at the same position and pulses of light are emitted from the top mirrors (designated as the white dots). In They both move at c from this point. (as indicated by expanding circle). Also note that the pulse moving off to the right always stays between the mirrors of the light clock moving to right. Thus from the perspective of someone moving with this light clock, the pulse just go back and forth between the mirrors.

Each clock is designed so that 1 "tick" is 1 round trip between the mirrors. As can be seen, the "moving clock ticks once for every two ticks of the "stationary" one.

Now here is where the "time dilation" part comes in. The speed of light is invariant. This means everyone measures it as having the same speed relative to themselves.
Thus if the "stationary" clock measures 1 nano second for the light to bounce back and forth between his two mirrors, The "moving clock" also measures 1 nano second between ticks of his clock (remember, from his perspective the light is just bouncing straight back and forth between his two mirrors at c). This is represented by the changing red numbers by each light clock which measure Time for each clock. The moving clock only counts to 1 in the time it takes for the stationary clock to count to 2. Thus from the stationary clock's clocks perspective, time runs 1/2 as fast for the moving clock.

 

[Stephanus]

Maybe the above animation will help.

Wow, that's a good animation and very clear, too.

 

[Stephanus]

Perhaps this simple diagram will help.
Let's forget about relativity, let's forget about time dilation.
Let's do phytagoras instead.

A train, its wide is 1 unit, its speed is 0.75 unit, it travel from A to B, it shines a light at A where/when it starts traveling. And let's forget about that, too.
Let's analyze the triangle, first.

This a basic pythagoras theorem. Can you figure out what AD length is? Remember, AF and DF is perpendicular.

 

[Stephanus]

The two postulates of special relativity are covariance (inertial frames have the same physics) and universality of c regardless of relative motion.

Yes, that's right.
I'm new in SR, too. But I know what I know, so I won't mislead you.
2. "and universality of c regardless of relative motion."

A. A fighter aircraft flies at 1000 kmh. It fires a sidewinder missle which has 3000kmh speed.
The speed of the sidewinder as seen by
The pilot:
3000 kmh
A ground observer:
4000 kmh
-----------
3000kmh ≠ 4000 kmh
Actually it's not 4000 kmh, but something like $\frac{3000+4000}{1+\frac{3000*4000}{c^2}}$ c is 300000 * 3600

B. A rocket travels at 250 thousands km/second. It shines a light which has 300 thousands km/second speed.
The speed of the light as seen by
The astronout:
300 thousands /sec
A ground observer:
300 thousands /sec, not 550 thousands!
---------------------------
300 000km/s = 300 000 km/s

Okay, this is first. But very important, can you understand this INTUITIVELY? Don't worry. I couldn't before, but I can, now

 

[Raman Choudhary]

Wow, that's a good animation and very clear, too.

Maybe the above animation will help. The two light clocks start at the same position and pulses of light are emitted from the top mirrors (designated as the white dots). In They both move at c from this point. (as indicated by expanding circle). Also note that the pulse moving off to the right always stays between the mirrors of the light clock moving to right. Thus from the perspective of someone moving with this light clock, the pulse just go back and forth between the mirrors.

Each clock is designed so that 1 "tick" is 1 round trip between the mirrors. As can be seen, the "moving clock ticks once for every two ticks of the "stationary" one.

Now here is where the "time dilation" part comes in. The speed of light is invariant. This means everyone measures it as having the same speed relative to themselves.
Thus if the "stationary" clock measures 1 nano second for the light to bounce back and forth between his two mirrors, The "moving clock" also measures 1 nano second between ticks of his clock (remember, from his perspective the light is just bouncing straight back and forth between his two mirrors at c). This is represented by the changing red numbers by each light clock which measure Time for each clock. The moving clock only counts to 1 in the time it takes for the stationary clock to count to 2. Thus from the stationary clock's clocks perspective, time runs 1/2 as fast for the moving clock.

Yes, that's right.
I'm new in SR, too. But I know what I know, so I won't mislead you.
2. "and universality of c regardless of relative motion."
yes completely
A. A fighter aircraft flies at 1000 kmh. It fires a sidewinder missle which has 3000kmh speed.
The speed of the sidewinder as seen by
The pilot:
3000 kmh
A ground observer:
4000 kmh
-----------
3000kmh ≠ 4000 kmh
Actually it's not 4000 kmh, but something like $\frac{3000+4000}{1+\frac{3000*4000}{c^2}}$ c is 300000 * 3600

B. A rocket travels at 250 thousands km/second. It shines a light which has 300 thousands km/second speed.
The speed of the light as seen by
The astronout:
300 thousands /sec
A ground observer:
300 thousands /sec, not 550 thousands!
---------------------------
300 000km/s = 300 000 km/s

Okay, this is first. But very important, can you understand this INTUITIVELY? Don't worry. I couldn't before, but I can, now

 

[Raman Choudhary]

You've got this garbled. The light clock is moving with respect to observer B; that means that, to observer B, the light clock appears time dilated. It does not mean that observer B appears time dilated with respect to himself.
This looks garbled too. The light clock is ticking slower than observer B's clock (relative to observer B), because there are six ticks of observer B's clock for every two ticks of the light clock.

SIr i know this and I haven't got this garbled, i think you understand time dilation so well that you are unable to understand my question so I will rephrase it for the last time because this is the last set of words I have to express my question and I thank you for your patience, the following is my question:
Suppose the observor A(the one w.r.t whom that light beam clock is at rest) has been provided with a Digital clock and the observor B(w.r.t whom the light beam clock moves at v=c/2) has also been provided an identical Digital clock.
Suppose the length of the train compartment is c(3*10^8 meters).
Now according to A, he sees the photon going to the top and coming back to the bottom(with velocity c) of compartment after hitting the top. He looks at his clock when it has returned and the digital clock clearly shows 2 sec.
Now according to B, he sees the same photon going towards the top(with velocity c) but not perpendicular to the bottom of compartment and rather at a certain angle(we can calculate this angle easily),clearly we can see that to reach the top the photon has to travel a bigger distance according to B and lesser distance according to A.
Now comes my question sir,
B also has a digital clock which is identical to that of A's, so when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec and it's just that according to him(B) photon will take somewhat more time to reach the top as the distance it has to travel is more as compared to A's.So what I am saying here is that both the clock's show same time at every instant(i'm saying on the basis that both are identical) but it's just that in B's frame of reference somewhat more time will be taken due to the increment in the distance the photon has to travel to reach to the top.
So by my argument when A's clock shows 2 sec. B's clock too reads 2 sec. and it's just that the photon for A by this time has completed his 1 round journey whereas for B it's yet to reach the bottom and when both clocks show 3.40 seconds(2*gamma) then B sees the photon hitting the bottom .
So my conclusion of the whole eg. is :
The increment in length in B's frame has just caused the total time interval to differ although the rate of digital clock ticking is same.

 

[Stephanus]

[..]Now according to A, he sees the photon going to the top and coming back to the bottom(with velocity c) of compartment after hitting the top. He looks at his clock when it has returned and the digital clock clearly shows 2 sec.
Now according to B, he sees the same photon going towards the top(with velocity c) but not perpendicular to the bottom of compartment and rather at a certain angle(we can calculate this angle easily),clearly we can see that to reach the top the photon has to travel a bigger distance according to B and lesser distance according to A.
Now comes my question sir,
B also has a digital clock which is identical to that of A's, so when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec and it's just that according to him(B) photon will take somewhat more time to reach the top as the distance it has to travel is more as compared to A's.

No!. B clock's does not read 1!
B's clock will read...

(w.r.t whom the light beam clock moves at v=c/2)

$t=\sqrt{1^2 + (\frac{1}{2})^2} = \sqrt{1.25} = 1.118$
Because...

B will see the light travels at 1.118 * 300000 = 335,410.1966 km. So, for B, the light won't take 1 second to reach the top, but 1.118 sec.
For A? Of course 1 second.

Now according to B, he sees the same photon going towards the top(with velocity c) but not perpendicular to the bottom of compartment and rather at a certain angle(we can calculate this angle easily),clearly we can see that to reach the top the photon has to travel a bigger distance according to B and lesser distance according to A.

Now you should stop right here!

Now comes my question sir,

STOP
Longer distance according to B, shorter distance according to A, and if their clock show the same, than for B the light travels more than c.
Remember:

The two postulates of special relativity are covariance (inertial frames have the same physics) and universality of c regardless of relative motion.

So the clock couldn't be the same. Only the velocity is invariant!

 

[PeterDonis]

when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec

Ah, I see; what you're missing isn't time dilation, it's relativity of simultaneity. B's clock is identical in construction to A's, but it is not at rest relative to A's, so "the instant when according to A the photon has hit the top" only has meaning for A, not for B. In other words, the word "when" here means "when, according to A's rest frame", which is only relevant when talking about the tick rate of A's clock, not B's clock. It does not mean "when" in any absolute sense, because there's no such thing in relativity. Your argument assumes that there is, so your argument is based on a false premise.

what I am saying here is that both the clock's show same time at every instant(i'm saying on the basis that both are identical) but it's just that in B's frame of reference somewhat more time will be taken due to the increment in the distance the photon has to travel to reach to the top.

This can't be right, because it amounts to saying that B's clock does not read off time in B's frame of reference. The speed of the photon is the same in all reference frames, so the photon has to take more time to travel a longer distance. But that means B's clock must read that longer time; otherwise B's clock is not a valid clock, because a valid clock always reads off time in its rest frame. B's clock must read off time in B's rest frame; it can't read off time in A's rest frame.

Your error, again, is in the statement "both clocks show same time at every instant". "Every instant" has no absolute meaning; it is different in each frame of reference, so "every instant" according to A is different from "every instant" according to B.

 

[universal_101]

Now comes my question sir,
B also has a digital clock which is identical to that of A's, so when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec and it's just that according to him(B) photon will take somewhat more time to reach the top as the distance it has to travel is more as compared to A's.So what I am saying here is that both the clock's show same time at every instant(i'm saying on the basis that both are identical) but it's just that in B's frame of reference somewhat more time will be taken due to the increment in the distance the photon has to travel to reach to the top.
So by my argument when A's clock shows 2 sec. B's clock too reads 2 sec. and it's just that the photon for A by this time has completed his 1 round journey whereas for B it's yet to reach the bottom and when both clocks show 3.40 seconds(2*gamma) then B sees the photon hitting the bottom .
So my conclusion of the whole eg. is :
The increment in length in B's frame has just caused the total time interval to differ although the rate of digital clock ticking is same.

I really like your view on this, for it is very natural to think this way, OK let me try to show the differences between the two views(of yours and others).

Your view of the scenario is based on the absolute 'time' and variable 'speed of light', that's how you can easily explain the observation by 'A' and 'B', whereas similarly, others have the view of variable 'time' and absolute 'speed of light'.

You can easily realize that the difference is because of different interpretation, your version of explanation says the time rate is same for the two observers, it's just that light has to travel a longer distance in frame of B(which implicitly means different light speed for different observers). While the other view is about the light speed being same for the two observers, and implying that the time rate itself changes(that is instead of rate of time being same, the instant of light beam striking the ceiling or the floor is same in both frames, which means the light journey in two frames represent the ticking clock, implying a different time rate).

 

[Stephanus]

Your view of the scenario is based on the absolute 'time' and variable 'speed of light', that's how you can easily explain the observation by 'A' and 'B', whereas similarly, others have the view of variable 'time' and absolute 'speed of light'.

Now this one I like

 

[georgir]

Now comes my question sir,
B also has a digital clock which is identical to that of A's, so when A's clock shows 1 sec (the instant when according to A the photon has hit the top) B's clock must also read 1 sec...

And here you already messed up your reasoning. Digital or light or sand or biological, all clocks are affected by time dilation. They may be identical, but they are not moving identically so they will not show 1 sec at the same time.