# A question about measuring time via the speed of light

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1. Jun 8, 2014

### BiGyElLoWhAt

Let us have an observer moving at some velocity v.

And let our observer have 2 mirrors, one on either side of their vehicle, that they will use to measure time.

In this situation, our observer is using a single photon (or ray, it really doesn't matter) to measure time. A ray of light bounces from one mirror and hits the other. Our observer uses the formula $\frac{distance}{time}$ to calculate the speed of light. If the light indefinitely bounces back and forth between the same 2 spots in the (parallel) mirrors, then at rest, our observer will calculate a speed equal to c.

If our observer moves at a velocity v, they will calculate a value less than c.

This is because our observer's (with respect to an outside observer's) coordinate system is in motion, and thus he/she see's only one component of the lights velocity, lets assume the x component (our observer is traveling along the y axis)

To our rider (our subjective observer), light is moving back and forth along some line (for simplicity y=0) indefinitely. To an outside observer, light is traveling with the same x component, but also a y-component that accounts for the difference in measured values of c.

Now, let us assume that our riders mirrors also have a way of emitting a photon (SPDC or whatever). So we have the same photon bouncing back and forth between the mirrors, and giving a measured value of less than c, which he/she compensates for via time dilation.
Since our (subjective) observer, has a way of emitting light, they take advantage of this fact, and emit a photon with a velocity perpendicular to the surface of the mirror (because in their rest frame, this should cause it to bounce back and forth between the same 2 spots on the mirrors, just like the first photon). To our observers dismay, it doesn't. If our observer is traveling at .5c, then the angle (with respect to the perspective of our subjective observer) is 45 degrees. If our observer then measures the distance traveled based on the angle and distance between the 2 mirrors, and then uses that to calculate c, it will appear that this particular photon is moving faster than c, whereas to an outside observer it is merely a photon oscillating along a line (say y=1) as long as there are mirrors there to reflect it.

If our observer uses photon 1 to calculate the time dilation, it will appear as though time is moving slower (greater delta t) and if he/she uses photon 2 to calculate time, it will appear as though time is moving faster (greater delta x).

This is entirely different to an outside observer who would see the y-component of photon 1 and be able to calculate the accepted value of c, and would see that photon 2 is just oscillating along a line (y=1) and would be able to calculate the accepted value of c using photon 2 as well!

Am I misunderstanding the concept of time dilation or what's the explanation for this?

2. Jun 8, 2014

### Staff: Mentor

No. Suppose the two mirrors are one light-second apart. The light makes it from one mirror to the other in one second according to the rider's measurement. He calculates distance traveled divided by time to travel, and comes up with c.

The outside observer notes that the light travels a distance $\sqrt{1+v^2}$ light-seconds (I got this by drawing the slant-wise path of the light and using the Pythagorean Theorem). He also notes that the travel time was $\sqrt{1+v^2}$ seconds (I got this by Lorentz-transforming from the rider's coordinates to the outside observer's coordinates). He calculates the speed to be the distance traveled divided by the time to travel, and comes up with the same answer: c.

The x components are not the same. Using the rider's coordinates, the x component of the light's velocity vector is 1 and the y component is 0. Using the outside observer's coordinates, the x component of the light's velocity vector is $\sqrt{1-v^2}$ and the y component is v.

Be careful here. The angles are not the same for the two observers, just as the y-axis distance the mirrors move is not the same.

3. Jun 8, 2014

### BiGyElLoWhAt

So will the 2 photons not differ in the amount of time it takes to get from one mirror to the other?

If so, will photon 2 not arrive faster than photon 1 (the photon oscillating between the 2 mirrors wrt to our rider)?

If not, why not?

I will draw a diagram to try to explain this situation better, and maybe it can be of use in explaining to me what's going on here assuming that I am missing something key.

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• ###### relativity light clock.png
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Last edited: Jun 8, 2014
4. Jun 8, 2014

### BiGyElLoWhAt

perhaps case 2 could have been illustrated a little better, our observer sees the velocity of the mirrors.

So here is my train of thought with this, if we calculate the speed of light using photon 1 and it's path with respect to our riders perspective, we will have $\frac{d}{t}$ which using a stop watch in motion with our rider, will come out less than c. Our rider (being versed in GR and SR) decides to rewrite his speed formula to $\gamma \frac{d}{t} = c$ and solve for the time dilation constant.

Now lets calculate the speed of light using photon with respect to our observers perspective:
$c = \frac{d}{t}$ and this comes out exactly c, as this photon (as previously pointed out) has a velocity $<\sqrt{1-v^2}, v>$ with respect to our observer.

Our rider sees the velocity of photon 1 as $<\sqrt{1-v^2},0>$ and says "This must be equal to c!" and adjusts his time measurements accordingly.

now photon 2 comes into the picture. Our observer measures the velocity of photon 2 and gathers $<c,0>$, no problems there.

Our rider then measures the speed of photon 2 and gathers $<c , -v>$, and our observer just calculated the dilation of time using photon 1, he got that he needs to measure less time intervals (or make them bigger) to make the velocity of light equal to c. Now using photon 2 he sees he needs to measure more time intervals to make it equal to c (or shrink them).

Am I approaching this conceptually wrong; because this is leaving both our rider and myself scratching our heads.

Last edited: Jun 8, 2014
5. Jun 8, 2014

### stevendaryl

Staff Emeritus
Why do you think that? His stop watch will show the same time-dilation as the light clock. So the rider will measure speed c.

6. Jun 8, 2014

### BiGyElLoWhAt

Ok, I see your point with that.

I guess what I'm thinking is this:

Let's alter our situation slightly. Our rider and our observer have a way to communicate, and our observer is now running the stop watch. Our rider measures the distance, and our observer is measuring the time, giving the time to our rider, and our rider makes the calculations.

Is this applicable? Or is this the inconsistency that's causing this theory to break down in my mind?

7. Jun 8, 2014

### stevendaryl

Staff Emeritus
For the light pulse traveling up and down inside the car, if L is the distance between mirrors,

• From the point of view of the rider, the distance traveled by the light going from one mirror to the other is $L$, and the time is $L/c$
• From the point of view of the observer outside the car, the distance traveled by the light is $\gamma L$, and the time is $\gamma L/c$

They both measure speed c, but they measure different lengths and different times.

8. Jun 8, 2014

### Staff: Mentor

That doesn't make your case 2 any clearer; I don't understand what it means. It seems like you are saying that the rider, who is moving along with the mirrors (i.e., the mirrors are at rest relative to him) can somehow "see the motion of the mirrors" in case 2. That's not possible. If the rider is moving along with the mirrors, he will see any photons moving back and forth between the mirrors as you have shown it in case 1. The only way the rider could see photons moving like your case 2 is if they were not bouncing back and forth between the mirrors, but were emitted by something else and just happened to pass through the rider's mirror assembly.

9. Jun 8, 2014

### Staff: Mentor

The distance in one frame over the time in another frame is not a speed in any frame.

10. Jun 8, 2014

### BiGyElLoWhAt

If you look, they are labeled, the left column is the riders perspective, the right column is our objective observers perspective.

Also case 1 and case 2 are the same situation, just separated by some time interval. Photon 2 bounces back and forth between the mirrors, makes its way to the edge of the mirror and shoots off into free space.

Last edited: Jun 8, 2014
11. Jun 8, 2014

### BiGyElLoWhAt

Yes. I see this, I didn't realize that was what I was actually doing until stevendaryl's post. I'm going to have to rethink things. Thanks everyone for helping clear this up.