I Does time dilation cause the speed of light to be invariant?

[Sentosa]

I'm trying to understand why the speed of light is the same for all observers. I have found different answers on-line. This page claims that it relates to time dilation.

But consider the following thought experiment: two ships flying at 98% c. Ship A is moving toward the sun, and ship B is moving away from the sun. The moment the two ships pass each other they are in the same gravitational field and are flying at the same speed. Time dilation should be the same for both. It would seem that light would be traveling at different speeds for the two ships, but is nevertheless measured to be the same speed. In this situation, it doesn't seem that differences in time dilation provide the explanation.

 

[Ibix]

You need to invoke time dilation, length contraction and the relativity of simultaneity to completely explain the invariance of the speed of light. You measure speed by measuring the distance traveled in a certain time. Each ship sees the other's clocks running slow, their rulers contracted, and notes that their clocks aren't correctly synchronised. These effects conspire so that each one will agree that the other's measurements lead to the same speed of light. But the two ships will give different reasons for why the measurements were what they were.

More typically, you assert the existence of an invariant speed and use that to derive relativistic effects (edit: you might want to look up the light clock or Einstein's train). The above is just running that backwards.

Leave gravity out of this. Newton's theory can't work in a relativistic universe and you need a lot of maths to handle Einstein's theory of gravity.

 

[alw34]

I'm trying to understand why the speed of light is the same for all observers.

That was also Einstein's dilemma. In the early 1900's some of the most famous scientists were struggling with Maxwell's theory of electromagnetism trying to understand why it conflicted with the 'standard' of the day, Newtonian mechanics.
The story is outlined here: https://en.wikipedia.org/wiki/Speci...2C_coordinates_and_the_Lorentz_transformation

In the Lorentz formulas shown there, t' and x' shown are stating how time and distance vary with relative speed! Despite what everybody "knew" was right back then, it turns out space and time were NOT the constants everybody thought, it was the speed of light that Einstein postulated was the REAL constant.

If you and I are in motion with respect to each other, our measures of time, and our views of each others distances are different. In fact, we don't agree on exactly when something happens. Two events in different locations occurring at the 'same time' is not absolute, it depends on the observers reference frame, the relative speeds of the different observers. This is what the first line of IBIX above post means.

 

[Dale]

I'm trying to understand why the speed of light is the same for all observers.

This is a hard question to answer because it is not really clear what you are asking and what you are willing to take as a given

For example, you could be convinced that there is an invariant speed and be asking why light happens to travel at that invariant speed.

Or you could be convinced that Maxwells equations describe light and be asking how we go from that to the idea that the speed of light is invariant.

Or... What is your foundation? What do you already understand/accept?

 

[Sentosa]

Or you could be convinced that Maxwells equations describe light and be asking how we go from that to the idea that the speed of light is invariant.

Or... What is your foundation? What do you already understand/accept?

I'm not familiar with Maxwell's equations. I understand velocity vectors in classical mechanics, which apparently don't apply in relativity. So why does light behave differently, from say, a tennis ball thrown from a moving bicycle?

 

[PeterDonis]

why does light behave differently, from say, a tennis ball thrown from a moving bicycle?

You're basically asking why the universe obeys the laws of relativity instead of the laws of Newtonian mechanics. The only real answer to that question is "because it does".

If you want something more, something like a reasonable argument based on accepted premises, you need, as Dale said, to tell us what your accepted premises are. So far all you've really told us is what they aren't.

 

[Sentosa]

My physics education was in classical mechanics. More recently I have read Relativity: A Very Short Introduction. I understand the relativity of space and time, although I need to learn more about the relativity of simultaneity.

Is it possible that light does in fact travel at different speeds for different observers, but they measure the speed at the same rate, because, for example, maybe their measuring equipment has been slowed down or sped up, compared to the other observer, because of time dilation or other variable?

 

[PeterDonis]

Is it possible that light does in fact travel at different speeds for different observers, but they measure the speed at the same rate, because, for example, maybe their measuring equipment has been slowed down or sped up, compared to the other observer, because of time dilation or other variable?

What would be the difference? If the "difference" in the speed of light for different observers is not detectable by any physical measurement, in what sense is it a difference?

 

[Dale]

I'm not familiar with Maxwell's equations. I understand velocity vectors in classical mechanics, which apparently don't apply in relativity. So why does light behave differently, from say, a tennis ball thrown from a moving bicycle?

It is going to be difficult to describe light without Maxwell's equations. A tennis ball follows Newton's laws, and light follows Maxwell's equations. They are very different equations so they behave very differently.

However, it seems like your question may be less about tennis balls and light and more about velocity addition, which follows the same rule for everything:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html

 

[Dale]

I understand the relativity of space and time, although I need to learn more about the relativity of simultaneity.

That is good. If you understand length contraction and time dilation, then you almost have the Lorentz transform.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html

From the Lorentz transform then you can easily show that c is invariant as follows:
A pulse of light from the origin travels at c in some frame. This is described as a sphere of radius ct as follows:
$c^2 t^2 = x^2 + y^2 + z^2$
Then, use the Lorentz transform equations to change everything into a different frame and simplify. You will get:
$c^2 t'^2 = x'^2 + y'^2 + z'^2$

So it travels at c in both frames.

 

[FactChecker]

My physics education was in classical mechanics. More recently I have read Relativity: A Very Short Introduction. I understand the relativity of space and time, although I need to learn more about the relativity of simultaneity.

Yes. The two observers disagree on what events are simultaneous. So they can not agree on how to set widely separated clocks to the same time. That is the root cause of the disagreement on time and distance. It also makes them both measure the speed of light as the same relative to their own motion. That is also why each observer thinks that the other's distance has shrunk without causing a paradox.

There is a story about Einstein realizing the disagreement of simultaneity when he was on a hill and saw two widely separated clock towers in town. That was a breakthrough.

 

[stevendaryl]

Is it possible that light does in fact travel at different speeds for different observers, but they measure the speed at the same rate, because, for example, maybe their measuring equipment has been slowed down or sped up, compared to the other observer, because of time dilation or other variable?

Every inertial observer has his own notion of distance between objects that are at relative to him. Every inertial observer has his own notion of the time between events. Speed is defined by (distance traveled) \div (trip time). So every inertial observer has the same value for this ratio.

The interesting thing about relativity is that the two numbers--D = distance traveled and T = trip time--are not the same for all observers, even though the ratio is, in the case of a light signal.

 

[Chestermiller]

In my judgment, for whatever it is worth (probably not much because I'm not a relativity guy), all that has been discussed in this thread so far are the effects rather than the causes. In my judgment, the fundamental cause of all these phenomena is the unique geometry of 4D spacetime.

Chet

 

[Herman Trivilino]

Is it possible that light does in fact travel at different speeds for different observers, but they measure the speed at the same rate, because, for example, maybe their measuring equipment has been slowed down or sped up, compared to the other observer, because of time dilation or other variable?

You're almost there. You just have to accept the fact that there is no special frame of reference in which to judge whether other clocks run slow or are out of sync, or that meter sticks are shortened. These things are true in all reference frames, so that means there is no basis upon which to claim that light really travels at different speeds in different frames. All you have are meter sticks and clocks, and when you use them to measure speeds you find that there is a maximum possible speed. It follows then, that such a speed must be the same in all reference frames.

 

[Dale]

all that has been discussed in this thread so far are the effects rather than the causes. In my judgment, the fundamental cause of all these phenomena is the unique geometry of 4D spacetime

I agree with your idea expressed here, but I try to avoid the word "cause" in this context. To me, the right term is "imply". If A causes B then not only does A imply B, but also A occurs before B.

The spacetime geometry does imply the invariance of c, but since it doesn't come before or after then I would just say imply rather than cause.

 

[alw34]

So why does light behave differently, from say, a tennis ball thrown from a moving bicycle?

A tennis ball thrown from a moving bicycle is not quite the simple event you might picture. It is a low speed approximation. There is a direct description here:

https://en.wikipedia.org/wiki/Velocity-addition_formula#Galilean_relativity

 

[lowemack]

I have the same trouble with this.

If I am traveling at 0.5C towards the sun and measure the speed of light from it, we get told that it still measures C because of our time dilation and length contraction. But the same reasons can't explain why the speed of light from the Earth, which I am moving away from, still measures C.

 

[alw34]

But the same reasons can't explain why the speed of light from the Earth, which I am moving away from, still measures C.

Whether you move toward or away from a light source, light still measures the same old 'c'. What would change is the frequency of the light you observe; that is, the same color source will appear as a different color.

Depending on just what your post means, you may also be invoking general relativity, in which gravity may further alter appearances. But nevertheless, locally light is always measured at 'c'.

 

[stevendaryl]

I have the same trouble with this.

If I am traveling at 0.5C towards the sun and measure the speed of light from it, we get told that it still measures C because of our time dilation and length contraction. But the same reasons can't explain why the speed of light from the Earth, which I am moving away from, still measures C.

Yes, it certainly can. There are actually three "relativistic" effects that work together:

1. Relativity of simultaneity: Events that are simultaneous in one frame are not simultaneous in a second frame.
2. Length contraction: An object that has a constant length as measured in its own rest frame will have a shorter length when measured from another frame.
3. Time dilation: A clock at rest in one frame will be measured to be running slow by observers in another frame.

The combination of these three effects result in light having the same speed in every reference frame. Of course, the reasoning actually went the other way: Einstein started with the assumption that light has the same speed in every reference frame, and derived those three effects.

 

[jartsa]

I'm not familiar with Maxwell's equations. I understand velocity vectors in classical mechanics, which apparently don't apply in relativity. So why does light behave differently, from say, a tennis ball thrown from a moving bicycle?

When measuring the speed of light, a small error in clock synchronization causes a large error in measurement, because a small time difference is being measured, because it takes just a small amount of time for the light to travel the distance between the clocks.

When measuring the speed of a slow tennis ball, a small error in clock synchronization does not cause a large error in measurement, because a large time difference is being measured, because it takes a large amount of time for the slow tennis ball to travel the distance between the clocks.

Clock synchronization error and clock synchronization difference are not different in this regard. Small errors and small differences become important as speed increases.

 

[Herman Trivilino]

If I am traveling at 0.5C towards the sun and measure the speed of light from it, we get told that it still measures C because of our time dilation and length contraction.

Yes, but it's best to work through the details and convince yourself that it's true, rather than accept it on authority.

But the same reasons can't explain why the speed of light from the Earth, which I am moving away from, still measures C.

Why not? The same reasoning does apply. To an observer on Earth your clocks are running slow and your meter sticks are contracted, so you measure the same value for the speed of a light beam as he does, whether the beam runs towards or away from Earth. Or whether you move towards or away from Earth.

 

[Ibix]

I have the same trouble with this.

If I am traveling at 0.5C towards the sun and measure the speed of light from it, we get told that it still measures C because of our time dilation and length contraction. But the same reasons can't explain why the speed of light from the Earth, which I am moving away from, still measures C.

Why don't we do the maths? In some frame S we see a light pulse moving along in the x direction. Its position is $(x,t)=( ct,t)$. Now we use the Lorentz transform to determine the location of the light pulse measured in a frame S', moving at speed v in the x direction. That is $$\begin{eqnarray} x'&=&\gamma (x-vt) = \gamma (ct-vt)=\gamma (c-v)t \\
t'&=&\gamma \left(t-\frac { vx}{c^2}\right)=\gamma \left(t-\frac { vct}{c^2}\right)=\gamma \left(1-\frac { v}{c}\right)t
\end {eqnarray} $$Substituting the last expression for t' into the last expression for x' we get that $x'=ct'$ - in other words the speed is always c, independent of v. You can make v negative, so this is true whichever way you are traveling with respect to the light pulse (feel free to try $x=-ct$ if that makes you feel better).

Since the Lorentz transforms are the mathematical statement of length contraction, time dilation and the relativity of simultaneity, then we can say that yes, they can explain the constancy of the speed of light.

 

[jartsa]

Why not? The same reasoning does apply. To an observer on Earth your clocks are running slow and your meter sticks are contracted, so you measure the same value for the speed of a light beam as he does, whether the beam runs towards or away from Earth. Or whether you move towards or away from Earth.

Because when your measuring apparatus moves away from light at speed 10 m/s, some effects must cause an "error" of 10 m/s in the result of the measurement.

But when your measuring apparatus moves towards the light at speed 10 m/s, some effects must cause an "error" of -10 m/s in the result of the measurement.

Length contraction and time dilation are quite useless in causing the right "errors", because they:

1: don't care about direction
2: are too small at slow speeds
3: cancel out each other

 

[Herman Trivilino]

Because when your measuring apparatus moves away from light at speed 10 m/s, some effects must cause an "error" of 10 m/s in the result of the measurement.

You spoke of time dilation and length contraction as the basis for always measuring the same value for the speed of a light beam. That's a calculation involving the division of a distance by a time.

Now, you're instead speaking of adding or subtracting two different speeds.

 

[jartsa]

You spoke of time dilation and length contraction as the basis for always measuring the same value for the speed of a light beam. That's a calculation involving the division of a distance by a time.

Now, you're instead speaking of adding or subtracting two different speeds.

When we calculate how much time dilation and length contraction change the result of measurement of the speed of a light beam, we add or subtract some speeds... And the effect is zero.

We have been ignoring the effect of relativity of simultaneity on speed measurements, so let's consider it now:

speed measurement result without relativity of simultaneity - speed measurement result with relativity of simultaneity = the effect of relativity of simultaneity on speed measurement

the effect of relativity of simultaneity on speed measurement = -10 m/s when clocks are moving at speed 10 m/s towards the light beam

the effect of relativity of simultaneity on speed measurement = 10 m/s when clocks are moving at speed 10 m/s away from the light beamTime dilation and length contraction can not be the basis for always measuring the same value for the speed of a light beam, because:
1: they don't care about direction
2: they are too small at slow speeds
3: they cancel out each other
4: relativity of clock synchronization takes care of measuring devices always measuring the same value for the speed of a light beam

 

[lowemack]

I understand that C is constant, and using the Lorentz formulas you can work it out. What I have trouble grasping, and a few others on here also, is that when we travel towards a light source at 0.25C, we would expect to measure the light at 1.25C. When we ask why not we get told, "ah, it's because your clock is running slow". OK, I can get that, but when we move away from a light source at 0.25C and expect to measure the light at .75C, how can it be because of the same reason "your clock is running slow"?

People have have been talking about "different reference frames", but in this example I am in the same reference frame measuring a light source in front of me and one behind me.

 

[stevendaryl]

Time dilation and length contraction can not be the basis for always measuring the same value for the speed of a light beam, because:
1: they don't care about direction
2: they are too small at slow speeds
3: they cancel out each other
4: relativity of clock synchronization takes care of measuring devices always measuring the same value for the speed of a light beam

I don't understand your claim. I would say, rather, that all three effects work together: relativity of simultaneity, time dilation, length contraction.

Let me rewrite the Lorentz transformations so that relativity of simultaneity and length contraction show up more clearly:

• x' = \gamma (x-vt) (This would be x' = x-vt without length contraction)

• t' = \frac{1}{\gamma} t - \frac{v}{c^2} \gamma(x-vt) (The term \frac{1}{\gamma} t reflects the effect of time dilation, while the term -\frac{v}{c^2}\gamma (x-vt) reflects the effect of relativity of simultaneity. Without time dilation, it would be t' = t - \frac{v}{c^2} \gamma(x-vt))

Now, compute speed in the x',t' coordinate system:

\frac{\delta x'}{\delta t'} = \frac{\gamma (\delta x - v \delta t)}{\frac{1}{\gamma} \delta t - \frac{v}{c^2} \gamma(\delta x - v \delta t)}

Getting rid of length contraction and time dilation means setting \gamma to 1, which would give us:

\frac{\delta x'}{\delta t'} = \frac{\delta x - v \delta t}{\delta t - \frac{v}{c^2} (\delta x - v \delta t)}
= \frac{\frac{\delta x}{\delta t} - v}{1 - \frac{v}{c^2} (\frac{\delta x}{\delta t} - v)}

So in the special case in which the speed in the unprimed frame is c, we calculate the speed in the primed frame as:
c' = \frac{c - v}{1 - \frac{v}{c^2} (c - v)}

The right-hand side is not equal to c. So you don't get light to have speed c in the primed frame, if you leave out time dilation and length contraction.

 

[jartsa]

@stevendaryl

Well, could we say that at non-relativistic speeds relativity of simultaneity rules, and time dilation and length contraction don't matter too much?

-------------------

Now let me try to spin this so that I didn't make any wrong claim at all:

Without the relativity of simultaneity the combination of time dilation and length contraction does not do anything, and by that I mean that a speed measuring device moving towards light beam measures the speed of the light to be c + own speed.

 

[stevendaryl]

Now let me try to spin this so that I didn't make any wrong claim at all:

Without the relativity of simultaneity the combination of time dilation and length contraction does not do anything, and by that I mean that a speed measuring device moving towards light beam measures the speed of the light to be c + own speed.

Yes, at least to lowest-order, you would have: c' = c \pm v

 

[Herman Trivilino]

When we calculate how much time dilation and length contraction change the result of measurement of the speed of a light beam, we add or subtract some speeds... And the effect is zero.

We take the distance and divide it by the time and get the speed. There is no addition or subtraction.

Suppose you fix two "buoys" in space along the line joining Earth and Sun. You separate them by a distance of 300 meters and you note that it takes a time of 1.0 µs for a light beam to travel from one to the other, regardless of which way the light beam travels. Now you fly by them, at a speed of about 0.87c, moving along the line joining them. You note that it takes 0.5 µs for the light beam to travel the distance between the buoys, but you also note that the distance between the buoys is 150 meters.

You can repeat the same thought experiment traveling in the opposite direction and you get the same result. You can have the light beam move in either direction and you get the same result.

the effect of relativity of simultaneity on speed measurement = -10 m/s when clocks are moving at speed 10 m/s towards the light beam

the effect of relativity of simultaneity on speed measurement = 10 m/s when clocks are moving at speed 10 m/s away from the light beam

It doesn't work that way. If you think it does, try doing the actual math by calculating these effects at various different speeds. Instead, the way you do the math is with the expression $$\frac{u+v}{1+\frac{uv}{c^2}}.$$
Let $u$ equal $c$ and let $v$ equal $\pm 10 \ \mathrm{m/s}$. The value of the expression is always $c$ as long as either $u$ or $v$ equals $c$.

 

[Orodruin]

You note that it takes 0.5 µs for the light beam to travel the distance between the buoys, but you also note that the distance between the buoys is 150 meters.

No it doesn't. The buoys are moving. How long it takes depends on the direction.

 

[Ibix]

I understand that C is constant, and using the Lorentz formulas you can work it out. What I have trouble grasping, and a few others on here also, is that when we travel towards a light source at 0.25C, we would expect to measure the light at 1.25C. When we ask why not we get told, "ah, it's because your clock is running slow". OK, I can get that, but when we move away from a light source at 0.25C and expect to measure the light at .75C, how can it be because of the same reason "your clock is running slow"?

People have have been talking about "different reference frames", but in this example I am in the same reference frame measuring a light source in front of me and one behind me.

As I noted, you can run my calculation with a forward-propagating light beam (where $x=ct$) and a rearward propagating light beam (where $x=-ct$) and come up with the same results. In your frame at time t', the forward propagating light beam is at $x'=ct'$ and the rearward propagating beam is at $x'=-ct'$, which means both are traveling at speed c.

The easiest way to explain it is "that's the way it is" and, from there, derive length contraction, time dilation and the relativity of simultaneity. But if you want to do it "backwards" as it were, the explanation turns out to be a mix of the three effects - length contraction, time dilation and relativity of simultaneity. The last one tends to get the least press but it probably the most important in explaining why your intuition is wrong.

Basically, if I lay out two rulers end to end and trigger light pulses from the middle going in opposite directions, I expect them to arrive simultaneously at the ends. However, you see the rulers in motion. That means that one pulse is "chasing" the end of a ruler that is running away from it, while the other meets the end of the ruler coming the other way. But that's fine; the concept of simultaneity is relative, so you don't expect the pulses to arrive at the ends at the same time.

For a better explanation (it's been a long day - sorry) try searching for Einstein's train. There are plenty of videos and descriptions. It's basically the reverse of the scenario I described above - the light pulses come from front and back of a train and meet in the middle (or not).

 

[Herman Trivilino]

No it doesn't. The buoys are moving. How long it takes depends on the direction.

If the observer is moving along the line joining them, at a speed of about 0.87c, the distance between them will be contracted to 150 m. It will therefore have to take a light beam 0.5 µs to traverse that distance at speed c. Where am I going wrong?

 

[jbriggs444]

In the observer's rest frame, the light is moving either doing a stern chase on a receding object or is moving toward an oncoming object. In either case the closing velocity is not c.

 

[Orodruin]

If the observer is moving along the line joining them, at a speed of about 0.87c, the distance between them will be contracted to 150 m. It will therefore have to take a light beam 0.5 µs to traverse that distance at speed c. Where am I going wrong?

The buoys are moving in this system so the separation speed between a light pulse and a buoy in this system is between c-v and c+v. In fact, it is just an example of a light clock. The total round trip will actually take longer in the frame where the buoys are moving.

 

[Herman Trivilino]

In the observer's rest frame, the light is moving either doing a stern chase on a receding object or is moving toward an oncoming object. In either case the closing velocity is not c.

The buoys are moving in this system so the separation speed between a light pulse and a buoy in this system is between c-v and c+v. In fact, it is just an example of a light clock. The total round trip will actually take longer in the frame where the buoys are moving.

Ahhh ... I see my error. In the observer's rest frame the buoys are moving, the distance between the buoys is 150 m, but since the buoys are moving the distance traveled by the light beam is more than 150 m when the beam moves in the direction of the buoys' motion, less than 150 m when the beam moves in the direction opposite to the buoys' motion.

 

[Orodruin]

Ahhh ... I see my error. In the observer's rest frame the buoys are moving, the distance between the buoys is 150 m, but since the buoys are moving the distance traveled by the light beam is more than 150 m when the beam moves in the direction of the buoys' motion, less than 150 m when the beam moves in the direction opposite to the buoys' motion.

Right, and the round trip (going once back and forth) actually takes longer due to time dilation.

 

[lowemack]

If speed is simply distance traveled divided by time taken then, if I go to Proxima Centura in my space ship, I know it is 4.2 LY away, I could go fast enough that only 2.1 years pass on my clock on my space ship. Why could i not say I traveled at twice the speed of light?

 

[stevendaryl]

If speed is simply distance traveled divided by time taken then, if I go to Proxima Centura in my space ship, I know it is 4.2 LY away, I could go fast enough that only 2.1 years pass on my clock on my space ship. Why could i not say I traveled at twice the speed of light?

Good question. The statement "light always travels at speed c, and nothing can travel faster than c" only makes sense for a very special class of coordinate systems, the inertial coordinate systems. For more general coordinate systems, such as the noninertial coordinate system of an accelerating rocket, that statement is not true (without some additional clarifications). What you can say, which is always true, is that nothing can outrun light. If in your trip, you had sent a light signal from Earth to Proxima Centauri, then it would have reached its destination before you did.

 

[Orodruin]

I know it is 4.2 LY away

This statement is only true in a reference frame which does not have a significant velocity relative to the stars. It will not be true if you are moving at a relativistic speed relative to these systems.

Furthermore, it is not you who are traveling in your inertial frame.

 

[Ibix]

To expand slightly on Orodruin's response, the distance between the stars will be length contracted by more than a factor of two in a frame where your clocks show 2.1 years to arrive at Proxima. The result is that you won't see anything exceeding the speed of light.

The situation is rather more complicated if you wish to consider non-instantaneous acceleration, as stevendaryl appears to be doing. But, as he says, you will never outrun light however you travel.

 

[Orodruin]

in a frame where your clocks show 2.1 years to arrive at Proxima.

"In a frame where it takes Proxima 2.1 years to arrive to you".

 

[Ibix]

"In a frame where it takes Proxima 2.1 years to arrive to you".

Indeed. I was trying to phrase it in such a way as to be interpretable either way about that point (apparently unsuccessfully) because it didn't seem directly relevant to the question. I should probably just have bitten the bullet.

 

[Sentosa]

Sorry to resurrect my own thread, but I've been thinking about this issue some more. The great paradox in this, it seems to me, is that the key to understanding the invariance of the speed of light is that you have to understand the nature, not of light, but of space and time. It is the flexibility of space and time (distance and time) that causes the speed (a ratio of distance over time) of light to be the same for all observers. This leads me to the following thought experiment, which I can't understand correctly:

Suppose I'm traveling in a spaceship at 95% c. I have a meter stick and a clock, and I decide to measure the speed of light in the ship's lab. Travelling at near c, the clock slows down (time dilation), and the meter stick contracts (length contraction). However, in order to get the same ratio (300,000), if the ruler contracts, shouldn't the clock speed up?

 

[Dale]

Travelling at near c, the clock slows down (time dilation), and the meter stick contracts (length contraction). However, in order to get the same ratio (300,000), if the ruler contracts, shouldn't the clock speed up?

Don't forget the relativity of simultaneity.

 

[Ibix]

Suppose I'm traveling in a spaceship at 95% c. I have a meter stick and a clock, and I decide to measure the speed of light in the ship's lab. Travelling at near c, the clock slows down (time dilation), and the meter stick contracts (length contraction). However, in order to get 300,000 km/s (a ratio), if the ruler contracts, shouldn't the clock speed up?

What does you measuring light speed look like to me, at rest in the "zero velocity" frame you are implying? You have a sensor at each end of a rod, and each sensor triggers when the light pulse passes it. So one sensor triggers, then the other. But in the time between the sensors triggering, the rod has moved. So, from my perspective, you aren't measuring the time taken for the light pulse to traverse the length of the rod, but the time taken for the light pulse to traverse the rod plus some velocity dependent distance. Combining time dilation and the relativity of simultaneity I will have no trouble explaining your results.

To put some maths into this, use the Lorentz transforms. Start with a rod of length L lying parallel to the x axis. Send a light pulse along it starting at $(x,t)=(0,0)$ and finishing at $(x,t)=(L,L/c)$, boost them both to a frame with velocity v, and see what the resulting $\Delta x'$ and ''\Delta t'## are.

 

[peety]

If the lab is 300,000 km long and you are in the lab you will time 1 second regardless of whether you are moving. An outside observer, given the right set up, will see the lab 150,000 km long and time half a second. He will also see your rods halved in length, but you won't. See the Ehrenfest paradox where there is confusion about which rods have shrunk.

 

[jbriggs444]

If the lab is 300,000 km long and you are in the lab you will time 1 second regardless of whether you are moving. An outside observer, given the right set up, will see the lab 150,000 km long and time half a second. He will also see your rods halved in length, but you won't. See the Ehrenfest paradox where there is confusion about which rods have shrunk.

The outside observer will see the lab 150,000 km long, but will not see the time taken by a light pulse as one half second. That is because the outside observer will not see the light pulse traveling 150,000 km. The pulse will start at one end of the lab, but by the time it reaches the other end, the target end will have moved. One needs relativity of simultaneity to reconcile the results properly.

 

[Orodruin]

An outside observer, given the right set up, will see the lab 150,000 km long and time half a second.

This is not true, the lab is moving and this affects the timing. The time will only be halved for a round trip. You need to take the relativity of simultaneity into account.

 

[Vitro]

Suppose I'm traveling in a spaceship at 95% c. I have a meter stick and a clock, and I decide to measure the speed of light in the ship's lab. Travelling at near c, the clock slows down (time dilation), and the meter stick contracts (length contraction). However, in order to get the same ratio (300,000), if the ruler contracts, shouldn't the clock speed up?

Always remember that motion is relative. From your own perspective you are always at rest, never traveling. So the right way to think about it is that you are at rest in a spaceship and from your perspective your clocks are not dilated and your rulers are not contracted, they are just standard clocks and rulers. You measure the speed of light as distance / time and get the standard result c.

Are you asking what your experiment would look like from another observer's perspective who's moving relative to you? That's a different story. Always be aware which point of view (reference frame) you are assuming for your analysis and avoid accidental "frame jumping".