# Time dilation in horizontal clock

## Main Question or Discussion Point

hi, when talking about time dilation we base our idea(as i understand) such that; static observer relative to other one claims that lights' path is longer for him/her and since lights' speed is c for each of them time dilation occurs. This is quite obvious in vertical clocks but in horizontal clocks how this phenomenan occur?
We seem to accept the dilation in horizontal clocks depending on vertical clocks not by observing the distance light travelled, obviously i am missing something probably about simultaneity, may be i need some visualization in horizontal clock time dilation.

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one thing to add when light ray goes in the same direction with moving source we can say it takes may be some longer distance and we can explain the dilation but when they are in opposite in direction the same effect makes the distance shorter. Is this related about dilation in some way?

JesseM
In a horizontal clock you have to take into account length contraction in the direction of motion, when you do you find that the moving horizontal light clock is dilated by the same amount as a vertical clock moving at the same speed. It's easier to start with a vertical clock since there are some basic physical arguments as to why there shouldn't be any length contraction in the vertical direction perpendicular to the axis of motion, see this thread for a discussion.

In a horizontal clock you have to take into account length contraction in the direction of motion, when you do you find that the moving horizontal light clock is dilated by the same amount as a vertical clock moving at the same speed. It's easier to start with a vertical clock since there are some basic physical arguments as to why there shouldn't be any length contraction in the vertical direction perpendicular to the axis of motion, see this thread for a discussion.
now i understand the length contraction in the direction of motion and i understand the constantnes of vertical length but one thing bothers me... Dilation in horizontal clocks, how do we observe the light in same speed in horizontal clocks, i know the formula is so simple length contracts and for light to be in speed c the time must dilate according to the formula x=ct but i can't seem to visulaize that both lights' speed is constant and time dilates.:)

JesseM
now i understand the length contraction in the direction of motion and i understand the constantnes of vertical length but one thing bothers me... Dilation in horizontal clocks, how do we observe the light in same speed in horizontal clocks, i know the formula is so simple length contracts and for light to be in speed c the time must dilate according to the formula x=ct but i can't seem to visulaize that both lights' speed is constant and time dilates.:)
Well, suppose the distance between the mirrors in the clock's rest frame is L, so in the clock's rest frame the time for it to make a two-way trip is 2L/c. Then in the frame where it's moving at speed v, the distance between the mirrors is reduced to L/gamma, where gamma=1/squareroot(1 - v^2/c^2). So suppose the light leaves the back mirror when the back mirror is at position x=0 at t=0 in the observer's frame, meaning in this frame the front mirror is at positon x=L/gamma at t=0. Then the light has position as a function of time given by x(t) = c*t, while the front mirror has position as a function of time given by x(t) = v*t + L/gamma. So to find when the light catches up with the front mirror in this frame, set these two equations equal, so c*t = v*t + L/gamma, which can be rearranged as t=L/(gamma*(c-v)). Plugging this back into x(t)=c*t, the light must catch up with the front mirror at position x=cL/(gamma*(c-v)). So now when the light is reflected back from the front mirror, its new position as a function of time must be x(t)=-c*(t - L/(gamma*(c-v))) + cL/(gamma*(c-v))...you can see that if you plug in t=L/(gamma*(c-v)) into this equation, then the light will be at position x=cL/(gamma*(c-v)) as it should be, and that the light is moving at speed c in the -x direction. Meanwhile, the back mirror started at position x=0 at t=0 and is moving in the +x direction at speed v, so its position as a function of time is just x(t) = v*t. So to figure out when the light returns to the back mirror, set both sides equal:

v*t = -c*(t - L/(gamma*(c-v))) + cL/(gamma*(c-v))
t*(c+v) = 2cL/(gamma*(c-v))
t= 2cL/(gamma*(c-v)*(c+v)) = 2cL/(gamma*(c^2 - v^2))

Now, 1/gamma can be written as squareroot(c^2 - v^2)/c, so the above can be rewritten as:

t = 2L*squareroot(c^2 - v^2)/(c^2 - v^2) = 2L/squareroot(c^2 - v^2)

And since gamma = c/squareroot(c^2 - v^2), this means 1/squareroot(c^2 - v^2) = gamma/c, so the above can be rewritten as t = gamma*2L/c. This is the total time in the observer's frame for the light to make a round trip from the back mirror to the front mirror and return to the back mirror. The total round-trip time in the clock's own frame was just 2L/c, so this shows that the round-trip time in the observer's frame has been expanded by a factor of gamma, just as the time dilation equation predicts.

If you're worried about the one-way times (back-to-front and front-to-back) and how they can be equal in the clock's frame but unequal in the observer's frame, here you have to get into the relativity of simultaneity and how watches traveling along with the front and back mirror which are synchronized in the light clock's own frame would be out-of-sync in the observer's frame--you might want to take a look at the numerical example I gave in [post=1561633]this post[/post].

JesseM
Incidentally, an a slightly simpler derivation of the two-way time could be done with the concept of "closing speed"--when the light is heading from back to front, the light is traveling at c while the front mirror is heading in the same direction at v, so the distance between the two is closing at a rate of (c-v), and since the distance between mirrors is L/gamma the time to go from back to front is L/(gamma*(c-v)). Then when the light is heading from front to back, it's heading towards the back mirror at c and the mirror is approaching the light head-on at v, so the distance between the two is closing at a rate of (c+v), so the time from front to back is L/(gamma*(c+v)). Thus the total two-way time is:

L/(gamma*(c-v)) + L/(gamma*(c+v)) = [L(c+v) + L(c-v)]/(gamma*(c-v)*(c+v)) =
2Lc/(gamma*(c^2 - v^2)) = 2Lc^2/(gamma*c*(c^2 - v^2))

and since c^2/(c^2 - v^2) = 1/(1 - (v/c)^2) = gamma^2, this reduces to:
2L*gamma^2/(gamma*c) = gamma*2L/c, the same two-way time found in the previous post.

ghwellsjr
Gold Member
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robphy
Homework Helper
Gold Member
I made an animation of one cycle of a circular light clock here so you can see both the horizontal and vertical orientations and all angles in between at the same time:

If you want more explanation and more animations see these two posts:

https://www.physicsforums.com/showpost.php?p=3059029&postcount=78

https://www.physicsforums.com/showpost.php?p=3059104&postcount=79
Nice animations.
If you plot spacetime diagrams of your animations, you'd get something that looks like my avatar.
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