Some strange questions about light

[JesseM]

This thread made me confused,
I know that the speed of light is c, a constant. And I thought that light traveled this distance compared to a place in the universe that is 100% still. so if you hypothetically traveled at the speed of c, and then decelerated with c, you would then stand 100% still.
But here i hear that if you are traveling at any speed, like 30% of the speed of light, you would observe the light as c! :O
let's say that someone shoots a photon in one direction. The shooter stands still, but you follows it with 50% of the speed of light. Would both measure that the photon is moving at the speed of c away from you?!
This makes no sense at all, but this is what i understand from this and other similar threads.

If it doesn't make sense to you, keep in mind that each observer measures speed in terms of distance/time according to his own set of rulers and clocks which are at rest relative to himself, with the clocks synchronized in his frame. But in relativity, each observer sees other observer's rulers and clocks as giving distorted readings, since each observer measures moving rulers squashed along their direction of motion by a factor of $$\gamma = 1/\sqrt{1 - v^2/c^2}$$, and moving clocks to be slowed-down by a factor of $$\gamma$$ and also out-of-sync with one another.

But if the shooter waits 10 seconds. the light would be 300 000 * 10 kilometres away (ca), and the one who chases the light in 50% of c, travels for 10 sec, then light also is 300 000 * 10 kilomtres away.
Where is the light after ten sec? Light would be 1.5 times longer away in the second case, so it doesn't make sense at all.

Suppose each observer has a long ruler at rest with respect to himself, and at each mark on the ruler is attached a clock, with all the clocks synchronized in that observer's frame (it's important to note that different frames disagree on 'simultaneity' in relativity, so clocks which are synchronized in their own rest frame will appear out-of-sync in other frames--if the clocks are synchronized and have a separation of x in their own rest frame, then in another frame where they're moving at speed v along the axis between them, the back clock will be ahead of the front clock by a time of vx/c^2). Each observer measures the light beam's speed by noting the time t1 on the clock at the mark m1 on his ruler as the light beam passes that mark, then later noting the time t2 on the clock at a different mark m2 on his ruler as the light beam passes that mark, and then calculating the speed as (m2 - m1)/(t2 - t1), or distance/time.

Given all this, here is a little example I put together on another thread to show how two observers will both measure a light beam to have a speed of c using their own rulers and clocks:

Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case $$\gamma$$ is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long. At the front and back of the ruler are clocks which are synchronized in the ruler's rest frame; because of the relativity of simultaneity, this means that in my frame they are out-of-sync, with the front clock's time being behind the back clock's time by vx/c^2 = (0.6c)(50 light-seconds)/c^2 = 30 seconds.

Now, when the back end of the moving ruler is lined up with the 0-light-seconds mark of my own ruler (with my own ruler at rest relative to me), I set up a light flash at that position. Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment. 100 seconds later in my frame, the back end will have moved (100 seconds)*(0.6c) = 60 light-seconds along my ruler, and since the ruler is 40 light-seconds long in my frame, this means the front end will be lined up with the 100-light-seconds mark on my ruler. Since 100 seconds have passed, if the light beam is moving at c in my frame it must have moved 100 light-seconds in that time, so it will also be at the 100-light-seconds mark on my ruler, just having caught up with the front end of the moving ruler.

Since 100 seconds passed in my frame, this means 100/1.25 = 80 seconds have passed on the clocks at the front and back of the moving ruler. Since the clock at the back read 0 seconds when the flash was set off, it now reads 80 seconds; and since the clock at the front read -30 seconds, it now reads 50 seconds. And remember, the ruler was 50 light-seconds long in its own rest frame! So in its frame, where the clock at the front is synchronized with the clock at the back, the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds, so since the ruler is 50-light-seconds long, the beam must have been moving at 50 light-seconds/50 seconds = c as well! So you can see that everything works out--if I measure distances and times with rulers and clocks at rest in my frame, I conclude the light beam moved at 1 c, and if a moving observer measures distance and times with rulers and clocks at rest in his frame, he also concludes the same light beam moved at 1 c.

If you like, I can also explain the details of your situation involving one person who shoots a light beam out and another who is moving at 0.5c relative to the shooter in the direction of the light beam (as seen in the shooter's frame), but the basic idea would be pretty much the same.

Ok, here is another thing I have been thinking of:
let's say that a person is moving at 80% of c away from a point, and another person moves at 80% of c in the opposite direction. Then they move 160% of c away from each other, right? Or is this impossible too?

As measured by the rulers and clocks of the frame where they are both moving at 0.8c, the distance between them would increase at a rate of 1.6 light-years per year. But when each one measures the other's speed using their own rulers and clocks, they will not find that the other is moving away from them at 1.6c; you can use the formula for addition of relativistic velocities given here to show that each will measure the other to be moving at (0.8c + 0.8c)/(1 + 0.8^2) = (1.6c)/(1.64) = 0.9756c.

 

[disregardthat]

That's a long and informative post, thanks. but I only understood parts of it, some things I don't understand, maybe you could explain?


"Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case $$\gamma$$
is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long."

What does that sign mean, and why does it get 1.25?


"Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment."

Also this I didn't understand. IS the ruler rotating? is it just moving forward?, and what does it read that is 0 seconds? And why is the clock 30 seconds behind yours?


"since the clock at the front read -30 seconds, it now reads 50 seconds."

I didn't understand this, where did you get the fifty seconds from?


"the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds"

Wouldn't you have to shoot the light beam at the front of the ruler for the rest of it to pass it?



Well, I have another question too, so this can make sense to me. If an object is moving 0.5c, then the time goes 0.5 of the normal?

And at the last bit, is it taken as a factor that you can not observe something before the light from it reaches your eyes? Or do you think of it as what we see is happening excactly there and then. (like: a sound we hear is not produced the excact same time we hear it, the sound-waves have to move through the air to reach our ears first)
Does it matter which DIRECTION we move to slow down time?
Sorry for all the questions, but this is kind of foggy to me, yet so disturbingly interesting.

 

[JesseM]

That's a long and informative post, thanks. but I only understood parts of it, some things I don't understand, maybe you could explain?

Sure.

"Say there's a ruler that's 50 light-seconds long in its own rest frame, moving at 0.6c in my frame. In this case $$\gamma$$
is 1.25, so in my frame its length is 50/1.25 = 40 light seconds long."

What does that sign mean, and why does it get 1.25?

I don't know if you caught it, but I did explain a little what it meant in the first paragraph of that post:

But in relativity, each observer sees other observer's rulers and clocks as giving distorted readings, since each observer measures moving rulers squashed along their direction of motion by a factor of $$\gamma = 1/\sqrt{1 - v^2/c^2}$$, and moving clocks to be slowed-down by a factor of $$\gamma$$ and also out-of-sync with one another.

To explain in a little more detail, that sign is the greek symbol "gamma", and as mentioned above it's equal to $$1/\sqrt{1 - v^2/c^2}$$. It's given its own symbol because this factor appears in a number of equations in relativity, so it's useful to have a shorthand. For example, if an object is moving at speed v in my frame, and if its length in its direction of motion is L in its own rest frame, then in my frame its length along this axis is given by $$l = L*\sqrt{1 - v^2/c^2} = L/\gamma$$. And if a clock is moving at speed v in my frame, and between two ticks it elapses a time of T in its own rest frame, then in my frame it elapses a time of $$t = T/\sqrt{1 - v^2/c^2} = T*\gamma$$.

"Let's say at this moment the clock at the back of the moving ruler reads a time of 0 seconds, and since the clock at the front is always behind it by 30 seconds in my frame, then in my frame the clock at the front must read -30 seconds at that moment."

Also this I didn't understand. IS the ruler rotating? is it just moving forward?, and what does it read that is 0 seconds? And why is the clock 30 seconds behind yours?

This was something I had also given a quick explanation for earlier in the post when I said:

(it's important to note that different frames disagree on 'simultaneity' in relativity, so clocks which are synchronized in their own rest frame will appear out-of-sync in other frames--if the clocks are synchronized and have a separation of x in their own rest frame, then in another frame where they're moving at speed v along the axis between them, the back clock will be ahead of the front clock by a time of vx/c^2)

The 30-second time difference between the two clocks as seen in my frame is based on that equation vx/c^2 -- the two clocks are on either end of the ruler, and the ruler is 50 light-seconds (ls) long in its own rest frame, and in my frame it's moving at 0.6c, so if the clocks are synchronized in the ruler's frame then they must be out-of-sync by vx/c^2 = (0.6 ls/s)*(50 ls)/(1 ls/s)^2 = 30 s in my frame.

Again, the "relativity of simultaneity" is a basic feature of relativity--if two events happen simultaneously (at the same time-coordinate) in one frame, they will have happened at different times in all other frames. For example, if two clocks at different locations both tick 12:00 at the same time in the clocks' rest frame, so they're in sync in that frame, then in other frames they will have ticked 12:00 at different times, so they're out-of-sync.

The relativity of simultaneity can be understood as a consequence of the procedure for "synchronizing" clocks in relativity--according to what's called the "Einstein synchronization convention", clocks should be synchronized using light-signals, making the assumption that light travels at the same speed in all directions. If you make this assumption, then you can synchronize two clocks at rest in your frame by setting off a flash at the midpoint between them, and then setting them to read the same time at the moment the light from the flash reaches each one. But this procedure automatically leads to disagreements about simultaneity. Suppose I am on a ship which is moving forward at high speed in your frame, and I set off a flash at the midpoint of the ship to synchronize two clocks at the front and back. In your frame, the back of the ship is moving towards the point where the flash was set off, while the front of the ship is moving away from that point, so if you assume light travels at the same speed in all directions in your own frame, then you should say that the light will catch up with the clock at the back before it catches up with the clock at the front. So if I set the clocks to both read the same time when the light catches up with them, you will see the back clock being ahead of the front clock.

"since the clock at the front read -30 seconds, it now reads 50 seconds."

I didn't understand this, where did you get the fifty seconds from?

From the fact mentioned at the top of that paragraph that "80 seconds have passed on the clocks at the front and back of the moving ruler." The clock at the front read -30 seconds at the time the light was emitted at the back (in my frame), and the clock elapsed 80 seconds between that moment and the moment the light reached the front (again in my frame, although 100 seconds elapsed between these moments according to my own clocks), so when the light reached it the clock would read -30 + 80 = 50 seconds.

"the light flash was set off at the back when the clock there read 0 seconds, and the light beam passed the clock at the front when its time read 50 seconds"

Wouldn't you have to shoot the light beam at the front of the ruler for the rest of it to pass it?

You could just imagine a spherical flash sending light in all directions, but if you want to think of it in terms of a beam being sent in a particular direction, then yes, it would have to be aimed in the direction of the front. In my scenario my ruler and the moving ruler are moving in parallel, and the flash is set off at the point in space and time where the back of the moving ruler and the back of my ruler are lined up, so you can assume it's aimed in the direction of the fronts of both rulers. The whole problem can be assumed to be in one dimension, the other spatial directions aren't important here.

Well, I have another question too, so this can make sense to me. If an object is moving 0.5c, then the time goes 0.5 of the normal?

No, as I said at the beginning of that post, "and moving clocks to be slowed-down by a factor of $$\gamma$$", and with v = 0.5c, $$\gamma$$ would equal $$1/\sqrt{1 - 0.5^2}$$ = about 1.1547.

And at the last bit, is it taken as a factor that you can not observe something before the light from it reaches your eyes? Or do you think of it as what we see is happening excactly there and then. (like: a sound we hear is not produced the excact same time we hear it, the sound-waves have to move through the air to reach our ears first)

The usefulness of having rulers with multiple synchronized clocks at different points along their length is that they allow you to assign coordinates to events using only local measurements. For example, if I look through my telescope in 2006 and see a distant explosion, and I see that it happened next to a mark on my ruler which is 3 light-years from where I am, then I can look at the clock sitting on that mark and see that it read a date of 2003 at the moment the explosion happened, so that would be the time-coordinate I'd retroactively assign to the event, not the time that the light from the event actually reached me.

Does it matter which DIRECTION we move to slow down time?

No, any clock moving at speed v will be slowed down by a factor of $$1/\sqrt{1 - v^2/c^2}$$ in my frame, regardless of its direction (although note that in that clock's own frame, it will be my clock that's slowed down by this amount--time dilation is relative, there is no true answer to which clock is 'really' running slower as long as both are moving inertially at constant speed and direction).

Sorry for all the questions, but this is kind of foggy to me, yet so disturbingly interesting.

No problem, and keep asking questions as they come up, it's the best way to learn about this stuff.

 

[disregardthat]

I think I have understood this reasonable properly, thank you very much!

Although I did not understand fully the ruler example, but in your further explanations I did understand the concept. So I guess it wouldn't matter.

There is no excact time in the universe that is "universal", everything is "out-of-synch" to each other if it is moving at a different speed. So I guess that this means that there is no point or speed (if I could call it that) that is entirely still in the universe. Like a "still" point would be moving at 10 km\h if an observer observed the point moving at this speed. Although both would claim that they were standing "still". (No acceleration, only speed)

I hope I will learn more about this when I choose physics in school.

 

[rbj]

Usually in relativity physicists just talk about the "rest mass", which is constant; but there's a separate concept called "relativistic mass", which is gamma*rest mass, where "gamma" is given by the formula $$1/\sqrt{1 - v^2/c^2}$$. If v=c, then gamma is 1/0; but if the rest mass is also 0, as in the case of a photon, then this formula tells you the relativistic mass would be 0/0, an undefined quantity, which I think is what Michael879 was talking about. To find the actual relativistic mass of the photon, you have to use a formula from quantum mechanics relating momentum and wavelength.

what I've suggested is not thinking of the relativistic mass of a photon in terms of it's rest mass (or "invariant mass") like you do for slower particles such:

$$m = \frac{m_0}{ \sqrt{1 - \frac{v^2}{c^2}} }$$

that leads to the undefined 0/0 problem. instead, think of the rest mass in terms of the relativistic mass:

$$m_0 = m \sqrt{1 - \frac{v^2}{c^2}}$$

and with

$$E = m c^2 = h \nu$$

or

$$m = \frac{h \nu}{c^2}$$

the rest mass of a photon comes out to be

$$m_0 = \frac{h \nu}{c^2} \sqrt{1 - \frac{v^2}{c^2}}$$

which is zero because $v = c$ for a photon (although i have seen this fact also questioned). anyway, no 0/0 problem for that.

 

[disregardthat]

Can someone answer if there is a point in the universe that is entirely still, or is everything relative to each other. If you move at 300 000 km\s less than light, wouldn't you stand still then? as the universal "speed" for a position with no movement at all.

 

[JesseM]

Can someone answer if there is a point in the universe that is entirely still

No, in relativity there is no notion of absolute speed, so likewise no notion of absolute rest. All the laws of physics will work the same way in the frames of any two inertial observers.

If you move at 300 000 km\s less than light, wouldn't you stand still then? as the universal "speed" for a position with no movement at all.

300,000 km/s less than light according to whose rulers and clocks? No matter who you are, if you are moving inertially and you use clocks and rulers at rest relative to yourself, you will find that you are moving at 300,000 km/s less than light (ie your speed will be 0 km/s and light's will be 300,000 km/s).

 

[disregardthat]

No, any clock moving at speed v will be slowed down by a factor of $$1/\sqrt{1 - v^2/c^2}$$
factor in my frame, regardless of its direction (although note that in that clock's own frame, it will be my clock that's slowed down by this amount--time dilation is relative, there is no true answer to which clock is 'really' running slower as long as both are moving inertially at constant speed and direction).

How can this be correct? I just read the twin paradox, and realized that if time is going slower for one person, and is not for the other, it will age less. But when the moving twin returns, almost untouched by age, his twin will have aged much. But for the twin untouched by age time would have gone slower for the twin standing on Earth in HIS frame... Something there is not correct...

 

[jtbell]

This "twin paradox" has been discussed many many many many times in this forum. Use the forum search function to look for the word "twin", and you will find enough reading material to keep you occupied for a long time.

 

[anantchowdhary]

Hey now the photon HAS mass but mass-->0.Also relativity does NOT exist at the speed of light,Therefore relativistic equations aint applicable to PHotons!

Also as Energy of a photon=hf,the photon has 0 energy at rest as it duznt hav frequency at rest.Hence it ceases to exist and therefore rest(invariant) mass=0