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

[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".

 

[Battlemage!]

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.

I think this is an underrated post. "The Principle of Relativity."

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.

I would say look at Mister T's post. But as a fun diversion:

When moving toward it you SEE the sun's clocking moving FAST, not slow. This is tied to the Doppler effect. You can see this if you examine what the twins paradox actually see rather than what some ideal clocks posted at places in space would read if someone was next to them at each event. As the twin leaves, both see the other as having slow clocks, and as he returns, both see the other as having fast clocks, but due to length contraction and relativity is simultaneity, "when" each twin sees the other have a fast or slow clock varies, and in doing the calculations in that thread it was found that this is actually why the "moving" twin ages less. This isn't difficult to work out. Just choose your relative speed, calculate your length contraction that Earth sees and compare measurements they will make. A good example showing this is found here:

http://www.scientificamerican.com/article/how-does-relativity-theor/

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?

I would point you to Mister T's post (quoted above). All inertial frames are equally valid, which means for all intents and purposes you are AT REST, your meter stick is NOT contracted, there ISN'T any time dilation, and you measure c for the speed of light.