Time Dilation equation questions

[TestOneTwo]

hello. ok,

In a philosophy class I'm taking the instructor went over the proof for time dilation using the pythagorean theorem and the distance=timexvelocity equation to end up with the equation
http://hsc.csu.edu.au/physics/core/space/9_2_4/Image4.gif
Based on measuring the pulse of light from the bottom to the top of a moving object, like a train.

and I understood that.
But what happens if you are measuring the light reflection from the back of the train to the front and from the front to the back? from the back to the front, it will appear to have traveled farther to an outside observer, implying time slows. But from the front to the back light would have appeared to have traveled a shorter distance. Doesn't that imply time speeds up? and these are happening at the same time, so I'm not sure what to think. I may just be thinking about this in some wrong way.

Can anyone help in showing steps to a mathematical proof to this like the one above?
I'm thinking that Distance = CxT+VxT going forward and Distance = CxT-VxT going back.
c being speed of light, t being time, v being the velocity of the train.
Not sure where to go from there though or how that could end up being equal to the original equation. So any help would be appreciated.

 

[yuiop]

Hi,

Imagine you have a train of length 2L. if the time it takes for a signal sent from the middle of the train to the front is t, the signal has to travel a distance L plus the distance the train moves forward in the time t which is vt, so the total distance the signal travels is L+vt which is the same as ct. So ct = L+vt.

We have not taken length contraction of the train into account here but let's call the contaction factor y so the equation becomes ct = Ly +vt.

See if you can work out the distance the signal travels to go from the middle of the train to the back and then work out how long it would take the forward and backward signals to arrive at the front or back ;)

 

[TestOneTwo]

so the distance on the way back would be: Ly -vt
I'm not exactly sure what you were asking at that last question, but I think you're askign how long it would take the ends of the train?

from the back to the front it would be 2Ly +2vt
front to back would be 2Ly -2vt
would these both equal 2ct?

 

[yuiop]

so the distance on the way back would be: Ly -vt
I'm not exactly sure what you were asking at that last question, but I think you're askign how long it would take the ends of the train?

from the back to the front it would be 2Ly +2vt
front to back would be 2Ly -2vt
would these both equal 2ct?

That's right.

I'm not exactly sure what you were asking at that last question, but I think you're askign how long it would take the ends of the train?

Ok so from the middle,

ct = Ly + vt (forwards) and ct = Ly - vt (backwards).

You now have enough information to calculate how long a signal sent from the middle will take to catch up with the front and how long it will take a signal sent from the middle to reach the back. When you have done that you might discover an interesting thing for yourself about time in relativity.

 

[TestOneTwo]

I'm not sure where to go from there.
It seems it takes longer for the signal to reach the front.

The time it would take forwards would be: t= Ly / (c-v)
backwards would be: t= Ly / (c+v)

but the person on the train would see them taking the same amount of time.

am I solving for the right thing?

 

[yuiop]

I'm not sure where to go from there.
It seems it takes longer for the signal to reach the front.

The time it would take forwards would be: t= Ly / (c-v)
backwards would be: t= Ly / (c+v)

Nicely done :)

but the person on the train would see them taking the same amount of time.

That is the point. The people on the train (whether it is moving or not) say the signals arrive at both ends simultaneously, while people outside the train say the signals do not arrive at the two ends at the same time. In relativity what is simultaneous for one observer is not necessarily simultaneous to another observer. The meaning of "at the same time" is not universal and depends on your state of motion. Does that seem surprising to you?

 

[TestOneTwo]

ah yes of course. I have read about that before, I just was not thinking of it here. I was too focused the aspect of time dilation.
so is that all you can learn measuring this way?
is there a way to use this way of measurement to reach the same conclusion of time dilation that you can reach when measuring from the bottom to the top of the cart?
i.e. http://hsc.csu.edu.au/physics/core/space/9_2_4/Image4.gif

 

[teodorakis]

ah yes of course. I have read about that before, I just was not thinking of it here. I was too focused the aspect of time dilation.
so is that all you can learn measuring this way?
is there a way to use this way of measurement to reach the same conclusion of time dilation that you can reach when measuring from the bottom to the top of the cart?
i.e. http://hsc.csu.edu.au/physics/core/space/9_2_4/Image4.gif

Yeah, that's the thing i asked another thread, but my english isn't good enough to describe it may be, so i quote from this thread.
Thanks.

 

[yuiop]

ah yes of course. I have read about that before, I just was not thinking of it here. I was too focused the aspect of time dilation.
so is that all you can learn measuring this way?
is there a way to use this way of measurement to reach the same conclusion of time dilation that you can reach when measuring from the bottom to the top of the cart?
i.e. http://hsc.csu.edu.au/physics/core/space/9_2_4/Image4.gif

Yeah, that's the thing i asked another thread, but my english isn't good enough to describe it may be, so i quote from this thread.
Thanks.

Yep, you can work out time dilation from the train. Using the conclusions that testonetwo came to:

The time it would take forwards would be: t= L*y / (c-v)
backwards would be: t= L*y / (c+v)

.. the time for a signal to go forwards and return to the middle would be:

$$t = L*y+(c-v) + L*y/(c+v)$$

Simplifying the above gives:

$$t = Ly*2*c/(c^2-v^2)$$

We already know the factor we used for the length contraction (y) is $\sqrt{(1-v^2/c^2)}$ and substituting this into the equation above it gives:

$$t = \frac{2L}{c} \frac{1}{\sqrt{(1-v^2/c^2)}}$$

The time measured in the frame of the train (T) is simply:

$$T = \frac{2L}{c}$$

So the time dilation ratio is:

$$\frac{t}{T} = \frac{1}{\sqrt{(1-v^2/c^2)}}$$

Of course this is not a proper derivation of time dilation as we had to know or assume the length contraction factor in the first place. This is why it is much better to use the vertical light clock in the train to work out the time dilation as the length contraction assumption issue is avoided altogether.

 

[teodorakis]

Of course this is not a proper derivation of time dilation as we had to know or assume the length contraction factor in the first place. This is why it is much better to use the vertical light clock in the train to work out the time dilation as the length contraction assumption issue is avoided altogether.

Thanks, but i am still like I'm lost in a circle, we get the contraction factor from vertical clock experiment and use it for horizontal clock as you explained, but shouldn't we get it alone and not make any assumptions in the horizontal version?The only assumption we should make is the speed of light is constant for all observers?

 

[yuiop]

Thanks, but i am still like I'm lost in a circle, we get the contraction factor from vertical clock experiment and use it for horizontal clock as you explained, but shouldn't we get it alone and not make any assumptions in the horizontal version?The only assumption we should make is the speed of light is constant for all observers?

The Michaelson Morley Experiment is effectively a comparison of the vertical and horizontal light travel times. The time dilation factor has to be the same in either direction so when you work that out you realize the signals would still not arrive back at the centre at the same time which contradicts what was measured in the experiment. By assuming that the speed of light is constant, you would have to conclude there is some other factor involved and that other factor is length contraction in the horizontal direction only. Hope that makes some sort of sense.

You could in principle measure the length contraction of the train directly and independently of time dilation. Once you have done that it would be easy to devise an experiment to measure the time dilation.

 

[teodorakis]

what i finally get is in the vertical clock version time is dialted by lorentz factor, and for this to happen in horizontal the lenghts must shrink by the same ratio, i think if we workout the algebra we could find this raito in vetical clock version i mean the horizontal distance is smllaer in that vertical clock experiment too so in that experiment if we put two clocks ;one vertical one horizontal ,and do some algebra i think we should find the same dilation, god I'm too silly i think to understand even the basic algebra:)

 

[teodorakis]

Yep, you can work out time dilation from the train. Using the conclusions that testonetwo came to:

.. the time for a signal to go forwards and return to the middle would be:

$$t = L*y+(c-v) + L*y/(c+v)$$

Simplifying the above gives:

$$t = Ly*2*c/(c^2-v^2)$$

We already know the factor we used for the length contraction (y) is $\sqrt{(1-v^2/c^2)}$ and substituting this into the equation above it gives:

$$t = \frac{2L}{c} \frac{1}{\sqrt{(1-v^2/c^2)}}$$

The time measured in the frame of the train (T) is simply:

$$T = \frac{2L}{c}$$

So the time dilation ratio is:

$$\frac{t}{T} = \frac{1}{\sqrt{(1-v^2/c^2)}}$$

Of course this is not a proper derivation of time dilation as we had to know or assume the length contraction factor in the first place. This is why it is much better to use the vertical light clock in the train to work out the time dilation as the length contraction assumption issue is avoided altogether.

But should'nt we able to derive it without length contraction assumption, for the horizontal light clcok, and all kind of clock?

 

[teodorakis]

let me ask my question again can we derive the time dilation eq in horizontal light clocks without accepting the length contraction at first?