Do my calculations describe the mechanics behind time dilation and length contraction

  • #1

Main Question or Discussion Point

I have a light clock onboard a spacecraft moving past the earth parallel to an observer. The lightclock measures the time it takes for light waves to bounce between two mirriors. I have two sets of mirriors, one on the vertical axis perpendicular to the direction of travel and also a horizontal set, in line with the direction of travel.

c = 10 m/s (speed of light)
v = 6 m/s (velocity)
l = 4 m ( length between two mirrors)

Firstly I take a time measurement onboard the craft:

[itex]ts[/itex] = 2l/c = 0.8

Then from earth's reference frame I calculate the time by using the following two equations, for the vertical clock I use:


[itex]te[/itex] = [itex]\frac{ts}{√1-vv/cc}[/itex] (please excuse my writing vv/cc, I mean v2/c2 however when I enter [sup[/sup] inside the fraction text it donsent seem to work :P im totally new to all this, trying to work it out as I go along, any help would be most kind)

[itex]te[/itex] = 1 second

Then for the horizontal clock:

[itex]te[/itex] = [itex]\frac{2lc}{cc-vv}[/itex] (apologies again the bottom half should read c2-v2)

[itex]te[/itex] = 1.25 seconds

.....clearly there is something wrong here, both clocks shold agree on the time.

So I apply the lorentz transformation to l in the horizontal clock:

[itex]Lo[/itex] = the proper length (the length between the mirrors in their rest frame)

[itex]L[/itex] = [itex]Lo[/itex]√1-v2/c2

and end up with a value of 3.2 for l in the horizontal clock....

so I go back to [itex]te[/itex] = [itex]\frac{2lc}{cc-vv}[/itex] (apologies again the bottom half should read c2-v2)

this time with 3.2 as my value for l, and then I get:

[itex]te[/itex] = 1 second

This means the amount of distance the light covers between each set of mirrors is equal, so even though the length is contracted in one clock, the 'light distance' is equal, as it is in both the rest frame and the moving frame.

Now both clocks agree and i'm very happy :) .... I hope!

Does this adequately describe the mechanics and interplay involving the speed of light, length contraction and time dilation?
 

Answers and Replies

  • #2
PeterDonis
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You appear to have done the math correctly in the Earth frame. (Though I would not advise giving your units as m/s, since the actual speed of light is clearly *not* 10 m/s! The usual practice is to use units in which c = 1, so you would have v = 0.6. It should be apparent that the actual numerical value of l, the length between mirrors in the spacecraft frame, is irrelevant since it drops out of the analysis; but if it's easier for you to do the math, you could just say l = 0.4 in the same system of units as above.)

For doing superscripts/subscripts in LaTeX, just use the ^ and _ characters. So your first formula would be, in LaTex,

t_{e} = \frac{t_{s}}{\sqrt{1 - v^{2} / c^{2}}}

which would appear like this:

[tex]t_{e} = \frac{t_{s}}{\sqrt{1 - v^{2} / c^{2}}}
[/tex]

Technically, for single characters as superscripts/subscripts, you don't have to enclose them in brackets; so you could just write, for example, t_e or v^2. But if there are multiple characters in the superscript or subscript, you need to bracket them, for example like this:

t_{Earth}

which appears as

[tex]t_{Earth}
[/tex]

as opposed to the (wrong)

t_Earth

which appears as

[tex]t_Earth
[/tex]

i.e., *not* what you want!
 
  • #3


Ahh! thankyou so much :) I was messing around with it for a while :p
Can you confirm then that my conclusions about this mechanic being at the root of time dilation/length contraction are indeed correct?
 
  • #4
PeterDonis
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Can you confirm then that my conclusions about this mechanic being at the root of time dilation/length contraction are indeed correct?
I'm not sure what you mean by "this mechanic". Do you mean, how does the Earth observer account for the fact that both light clocks "tick" at the same rate, even though one is length contracted and one is not?

First, it's worth noting that your two equations for the time can be shown to be identical without doing the actual calculations:

[tex]t_e = \frac{t_s}{\sqrt{1 - v^2 / c^2}} = \frac{2 L_0}{c \sqrt{1 - v^2 / c^2}} = \frac{2 L_0 \sqrt{1 - v^2 / c^2}}{c \left( 1 - v^2 / c^2 \right)} = \frac{2 L c}{c^2 - v^2}[/tex]

Second, I think the above equations capture something that your verbal description left out, which is that, in the Earth frame, the mirrors are *moving*.

The vertical clock's light signals are perpendicular to its motion, so the only factor affecting its "tick rate" in the Earth frame is time dilation.

But the horizontal clock's light signals are moving in the same direction as the mirrors themselves, so they have to catch up to one mirror (the one ahead), but move back towards the other mirror (the one behind). If there were no length contraction, this last effect would cause the horizontal clock to tick *slower* (because the "catching up" to the mirror ahead adds more time than the fast closure with the mirror behind takes away). That's why the horizontal clock's time came out longer before you took length contraction into account.
 
  • #5


Ahh yes, thankyou for showing me that :)

I guess I'm looking for a defining statement to define the reasons behind length contraction, time dilation. I have been given examples such as the train and the tunnel. But the example with the Light clock seems to show the relationship between light, spacetime and relative observations. It clearly shows mathematically why length contraction and time dilation are required to preserve lights properties.

Can I move on from this topic safely knowing my described understanding of lc, td is complete? Is there anything I have left out/missed?
 
  • #6
PeterDonis
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But the example with the Light clock seems to show the relationship between light, spacetime and relative observations. It clearly shows mathematically why length contraction and time dilation are required to preserve lights properties.
Yes, this is true.

Is there anything I have left out/missed?
Not that I can see.
 
  • #7


Many thanks for your help :)
 
  • #8
441
8


Ahh yes, thankyou for showing me that :)

... It clearly shows mathematically why length contraction and time dilation are required to preserve lights properties.
No, it shows the results of light speed being constant!
 
  • #9


No, it shows the results of light speed being constant!
does this contradict my statement?
 
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  • #10
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[..] It clearly shows mathematically why length contraction and time dilation are required to preserve lights properties.
Not entirely, but we're discussing that already in the thread of which this is a spin-off.
https://www.physicsforums.com/showthread.php?t=625509
Can I move on from this topic safely knowing my described understanding of lc, td is complete? Is there anything I have left out/missed?
You seem to be comfortable enough with applying length contraction and time dilation to basic optical experiments. :smile:

However, I don't think that one ever has a "complete" understanding.
For a more complete understanding of the "mechanics" of length contraction you should consider physically more difficult cases, such as Bell's spaceship as discussed here:
https://www.physicsforums.com/showthread.php?t=624284
 
  • #11


No, it shows the results of light speed being constant!
Surely we can preserve the speed of light without having to apply the contraction, as in the horizontal clock, before we apply the contraction the time is 1.25 seconds, so the speed of light is still preserved in both clocks. However the 'light time' is not preserved, so the contraction is needed.dosent this show that preserving the speed of light is not the only consideration?
 
  • #12
PeterDonis
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Surely we can preserve the speed of light without having to apply the contraction, as in the horizontal clock, before we apply the contraction the time is 1.25 seconds, so the speed of light is still preserved in both clocks. However the 'light time' is not preserved, so the contraction is needed.dosent this show that preserving the speed of light is not the only consideration?
The problem with thinking about the horizontal clock case as due to two effects combined--the light travel time, *plus* the contraction--is that without the contraction, the light beams in the horizontal and vertical clocks would not meet up; if the horizontal and vertical clocks both started out their light beams at the same instant, the vertical beam would return *before* the horizontal one.

But that can't be right, because in the rest frame of the clocks, the two beams do return at the same instant, and that's a direct observable, so it must be the same in all frames. The most natural interpretation of this direct observable is that it shows that the speed of light is the same in all directions. Length contraction of the horizontal mirror, in a frame in which the mirrors are moving horizontally, is required to preserve this direct observable--i.e., it's required to preserve the fact that the speed of light is the same in all directions, which requires that the vertical and horizontal light beams remain in sync.
 
  • #13
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The problem with thinking about the horizontal clock case as due to two effects combined--the light travel time, *plus* the contraction [..] [obvious deduction omitted]. But that can't be right, because in the rest frame of the clocks, the two beams do return at the same instant, and that's a direct observable, so it must be the same in all frames. [..]
:uhh: The two beams surely return at the same instant independent of the motion of the system, and other considerations (as discussed in parallel) show that it almost certainly must be right. :smile:
As a matter of fact, it has been shown that if the conservation laws as well as Maxwell's laws (which include the light principle) hold, then the relativity principle automatically follows. And the rest is, as you also say next, interpretation.
 
  • #14
PeterDonis
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The two beams surely return at the same instant independent of the motion of the system
Yes, and that means that saying "we can preserve the speed of light without having to apply the contraction" is incorrect. Without including the contraction, the speed of light is not "preserved".
 
  • #15


After sitting down and thinking about it for a little while longer, I realised that the 2 clocks were giving me two different time readings for one cycle of bounce. Which one is correct? As harrylin has alreday explained to me, vertical contraction is not an option and if I were to use the horizontal clock's time, I would need to alter the vertical length. So now I know that the correct time is given by the vertical clock.

I check the time on an identical clock on earth in a stationary reference frame and it reads 1 second (a little over one complete bounce between both mirriors)

I then look up into the sky and see that the clock on the ship reads 0.8 seconds (exactly one complete bounce)

just before the clock was started the ship sent out a light beam, directly in front of it. On earth we will measure that the light beam had spent one second traveling and had theifore covered 10 meters. Wheras on the ship the light beam had been traveling for only 0.8 seconds and had theifore only covered 8 meters.

we must conclude that in order to preserve the speed of light for all reference frames, that in fact length on the horizontal axis has contracted to mean that now, 10 meters in the stationary reference frame is equivalent to 8 meters in the moving one. \0/
 
  • #16
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Yes, and that means that saying "we can preserve the speed of light without having to apply the contraction" is incorrect. Without including the contraction, the speed of light is not "preserved".
I understood it to mean that the speed of light is preserved (constant) in the "stationary frame" - there's nothing wrong with the second postulate (he replied there to "the speed of light being constant"). And merely preserving the speed of light between reference frames (invariance) can also be done with another factor, as we explained in the other thread. So, it depends on what he meant with it.
 
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  • #17
3,872
88


After sitting down and thinking about it for a little while longer, I realised that the 2 clocks were giving me two different time readings for one cycle of bounce. Which one is correct? As harrylin has alreday explained to me, vertical contraction is not an option and if I were to use the horizontal clock's time, I would need to alter the vertical length. So now I know that the correct time is given by the vertical clock.[..]/
Sorry but I don't know what you mean - there is no "correct time" that we can determine, except the times that we read on a reference clock - and those times don't have a "vertical" or "horizontal", the position (neglecting effects from gravity) has no effect, just as you already knew. :bugeye:
You correctly calculated earlier that the 2 clocks give you two identical time readings for one cycle of bounce.
So, something (but what?) put you on a wrong track...
 
  • #18


Apologies, my last post was poorly worded. I was refering to the times given by the vertical clock and the horizontal one, before I applied the lorentz contraction. I was kinda thinking out loud, following the logical steps towards applying the contraction.
 
  • #19
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Apologies, my last post was poorly worded. I was refering to the times given by the vertical clock and the horizontal one, before I applied the lorentz contraction. I was kinda thinking out loud, following the logical steps towards applying the contraction.
OK - then your post #15 is effectively the same as your post #1 - only now you added the conclusion that vertical length change is not an option, and you used it as starting point. Fine. :smile:

And as a matter of fact, you could also use time dilation as starting assumption; nowadays this has been satisfyingly demonstrated.
 
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