# Questions about the speed of light

1. Mar 7, 2006

i read an article in howstuffworks.com a while ago about the speed of light and how light always moves at the speed of light from all frames of reference, and the author used an example where a parked car turns on its lights and the light hits a person standing down the street at the speed of light, and then a car driving at 50 mph turns on its lights and the light still hits the person at the speed of light, instead of the speed of light + 50 mph(small difference, i know). then there were other examples, and it started to confuse me... i'll have to find the article again and paste a link to it... anyway, me and a friend were talking about light recently and i'm still having a hard time grasping the whole concept, so if anyone could answer a few questions for me, i'd appreciate it

1. my first question is about the concepts of "frame of reference" and speed, i just want to make sure i have the basics down: i'd be correct in saying speed is always relative, right? like if i was running on the ground at ten miles an hour and threw a ball in front of me at 20 miles an hour, from the earths frame of reference the ball is moving at 30 mph, from my frame of reference the ball is moving 20 mph and the earth is moving at 10 mph, and from the balls frame of reference i'm moving at 20 mph and the earth is moving at 30 mph. is everything i said correct, am i using the term "frame of reference" properly?

2. if everything i said is true, then what causes time to slow down/speed up depending on an objects speed? since from an objects own frame of reference it's constantly still and speed can only be measured from other frames of reference (the surface of the earth, the sun, etc), what is it that determines for whom time will "slow down"? i'm assuming it has something to do with the shift in momentum, whichever object accelerates/decelerates, is that it?

3. gonna make an ascii drawing:
........................---O-------------->..................................
.........................ball B............x...............................|>= flashlight
....................................ball A O.....................................

lets say you have a flashlight, and a ball(ball A) that isnt moving (from the flashlight's frame of reference), and another ball is flying towards the flashlight at 100 mph (ball B), and when ball B passes ball A at point X, the flashlight turns on... both balls are an equal distance from the light when it turns on and light always moves at the speed of light from all frames of reference, including ball A and ball B, so (this is where i get completely lost) the light hits both balls at the same time? i've asked a few people about this, and they answered "yes, because time slows down for ball B, so even though ball B will hit the light first, if you had a clock on both ball A and ball B both clocks would read that the light hit at the same time", but what if ball B were going in the other direction at the same speed? time would be still be slowed down except now the distance between the light and the ball have increased instead of decreased, so wouldnt it be hit much later then ball A in that case?

4. if a spaceship were flying away from the sun at half the speed of light then turned left 45 degrees, would it catch more sunlight if it wasnt moving but still facing away from the sun at the same angle? how would the sun apear to the people inside the ship, would it look like an oval?

so far i'm still having a tough time grasping this whole thing. these are the explanations i've come up with to explain it so far:

A. the way i see of "frames of reference" and relativity is wrong, and speed is a constant compared to some point in the universe, and that light always moves at the speed of light no matter what speed the source is traveling at. though this wouldn't really explain some of the examples i've seen...

B. the way objects stretch/shrink when passing other objects is relative to the speed the object is coming toward or moving away from the other objects frame of reference(the speed at which the distance is being closed between the two objects), rather then just its relative speed. like if a man was standing 100 feet from a highway and a car was driving down at 50 mph, the speed at which the car is closing the gap between itself and the man would kind of resemble the way y=x^3 looks on a graph, and it's from this shrinking/stretching that distance is calculated when dealing with objects that are moving. this would explain why the light would hit both objects at the same time in the example with the balls i gave above: the distance is relative to each frame of reference, though this explanation contradicts with some things i've read so i don't think this is it either...

C. when light comes out from a source(a star for example), it doesnt come out in a sphere but rather in a weird amobea-like shape, stretching in areas to hit moving objects, so that it hits all objects that were equidistant from the lights source at the same time. it's completely ridiculous and i don't believe it, but it's the only explanation that works with most of the examples i've seen so far :P

if anyone could shed any light on this (no pun intended) that'd be great, TIA!

2. Mar 7, 2006

### chroot

Staff Emeritus
Your post is quite long, and it appears that all of your difficulty really surrounds this single sentence, so I'll only deal with this one sentence. Once you understand the basics, we can move up to discussing more complete examples.

No one ever measures time slowing down in their own frame of reference. Imagine you're the captain of a starship capable of relativistic travel between stars.

Imagine that, once you're settled into a course to a distant star at 0.9c, you pull the blinds down on all your starship's windows. There's no way that you will be able to "measure" your own speed; this much is probably already familiar to you. Velocity (speed) is relative, so you can't say what your speed is without comparing your starship's motion with the motion of some other object.

Time is very much the same way. You will never notice your wristwatch ticking "slowly," nor will you be able to devise any other experiment aboard your starship which will indicate any kind of time dilation. All your pendulum clocks and oscillating atoms and so on will behave exactly as they would under any other circumstance.

However, if you re-open the blinds and listen to the radio broadcasts from your pals back on earth, you'll notice the wavelengths stretched. If they flash a laser light in your direction once a second (according to their watches), the flashes will appear to be much longer than one second apart (according to your watch). Similarly, if you flash a laser light back at them, they'll think your clock is running slowly.

In truth, you cannot say whose clock is running slowly, because the very measurement of time is relative. All you can say is that someone else's clock, who is in relative motion, appears to be running slowly.

This is all a consequence of the first postulate of relativity -- all physics appears the same in all inertial reference frames. Time dilation (and length contraction) only occur when you translate measurements of time or distance from one reference frame into another.

- Warren

3. Mar 7, 2006

### Janus

Staff Emeritus
Okay
This is going to take more time than I have right now, but basically, it is always the other guy's time that slows down. If I watch a clock speed by me, I see his clock running slow, but he sees my clock run slow.
Here's the thing, when you say that the flashlight turns on at the same time as B passes x, you have to say in which frame this is true, A's or B's. If in A's then the the light doe not turn on when B passes x according to B, and vice versa. This is known as the "Relativity of Simultaneity".
But one thing is sure, the light will not hit A and B at the same time according to either A or B. According to A, B is rushing towards the light and hits the light first. According to B, A is rushing away from the light and thus it takes longer for the light to reach it, and again, the Light hits B first.

4. Mar 8, 2006

ah ok, thanks for the replies, this is starting to make a bit more sense to me so whenever you see an object in motion, time for that object will appear to move slower from your own frame of reference, never faster? if so, then what happens when people take a flight on the concord (if you put an atomic clock on the concord and then check it after it lands from a flight, it will be behind by a little when compared to time on earth, right?... or is it the other way around)? from the frame of reference of a person that's standing on the earth, time will be moving slower aboard the concord, which explains why a clock on the concord would be a few seconds behind "earth time", but from the frame of reference of a person aboard the concord, wouldnt time on earth have to appear to speed up since the clock on earth will be ahead of the clock that's been aboard the concord? what is it that determines which clock will be ahead of the other?

5. Mar 8, 2006

### Staff: Mentor

Your question is closely related to the classic relativistic "Twin Paradox" which has been discussed many many times in this forum. Try using the forum search tool to look for "twin" and you'll have enough reading material to keep you busy for a week.

Also try the entry on the Twin Paradox in the Usenet Physics FAQ:

6. Mar 8, 2006

### chroot

Staff Emeritus
There are actually two components to the Concorde experiment you outlined (by the way, this experiment has been done; it is known as the Hafele-Keating experiment): time dilation due to relative velocity, and time dilation due to differences in gravitational "field strength."

http://hyperphysics.phy-astr.gsu.edu/HBASE/relativ/airtim.html

- Warren

7. Mar 8, 2006

8. Mar 10, 2006

### haushofer

I have also a question concerning the speed of light. Is the speed of light also the same for all observers in accelerating frames? I have a feeling that the fact that you can always locally adopt a flat metric has to indicate that it should, but I am not sure about this. Could anyone help me on this one?

9. Mar 10, 2006

### pervect

Staff Emeritus
The local speed of light, measured using the clocks and rods of a local observer, is the same everywhere in an accelerating frame.

The coordinate speed of light, measured using the coordinate system of an accelerating obserer, is NOT the same everywhere.

The local observer adopts a locally Minkowskian metric, and thus the speed of light is constant and equal to 'c' in his local coordinate system, using his local clocks and rulers.

The global observer does not have a Mikowskian metric, and the coordinate speed of light is not constant.

I hope this helps.

10. Mar 11, 2006

### haushofer

Ehm, could you explain a bit more what exactly the difference is between a local observer and a global observer? Does it have to do with the space-time region where the measurement is taken? I am very much aware of the fact that one can only make the metric flat in one point, but I don't see how to connect this with your post.

11. Mar 11, 2006

### pervect

Staff Emeritus
Let's see if I can guide your thinking a bit:

First question: What is the Lorentz interval of a path travelled by a light beam? (Assume the light beam is in a near vacuum,not any media).

a) positive
b) zero
c) negative
d) don't know

Second question: using the results of the first question, if you are given a metric, can you write down a differential equation describing the path of a light beam?

Second question - example). For a more definite example, consider the example of the metric associated with the exterior of a black hole - the Schwarzschild metric. What would the equation representing the radial propagation of light be in Schwarzschild coordinates? Would the rate of change of the Schwarzschild R coordinate with respect to the Schwarzschild t coordinate of this path be constant for a radially ingoing or outgoing light beam?

Third question: using this differential equation, and the above example, can you answer your own question?

Last edited: Mar 11, 2006
12. Mar 11, 2006

### kmarinas86

A local observer has one point of observation. What about a global observer? For instance, think of this global observer as someone who the universe on his computer. The processor of his computer has a clock which governs the coordinate time. Without it, casuality would not work. On the screen there is a grid of the curvature of space time. This is the simulation's coordinate system. The brighter the grid lines intersecting, the faster proper time for a object that is on that intersection who is not moving relative to it. That is to say, for an observer on a brighter intersection, his clock will run closer to the computer's clock. The brighter the lines the higher the gravitational potential. The gravitational field makes possible an accelerated frame of reference for a object that is basically stationary to the computer's coordinate system. Now, because of this, near the dark the lines we may naively assume that the light, shown as a moving yellow line on computer screen is moving slower where the grid lines are darker. However in the same way gravitational time dilation increases, the gravitational field increases in strength to that same proportion. Therefore, even according to the computers clock, the light will be moving at the same speed on the screen, whether it's over the bright lines or the dark ones. However, according to an object at rest with the computer's coordinate system, depending on the position, the speed of light would vary? So either that, or acceleration does not increase by the factor which governs gravitational time dilation?

This appears to be what I have been told about the gravity of black holes. But if this were the case, then the speed of light wouldn't really be constant after all from an object's perspective (who is motionless relative to the computers coordinate system)! Anybody?

Last edited: Mar 11, 2006
13. Mar 15, 2006

### haushofer

Well, that Lorentzinterval should be zero. And then a differential equation...We can parametrise the path, and then obtain the expression of the Lorentzinterval in terms of the parametrisation:

$$ds= \sqrt{g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda} } d\lambda = 0$$

For the radial infalling light, we have that the angular part of the line element is 0, so we get

$$ds^2 = (1-2m/r)dt^2 - \frac{1}{(1-2m/r)} dr^2 = 0$$

so it seems to me that

$$\frac{dr}{dt}=(1-2m/r)^2$$

So dr/dt is dependent of r. But I still don't see how I can make a conclusion from this about my question... :uhh:

Last edited: Mar 15, 2006
14. Mar 15, 2006

### pervect

Staff Emeritus
If you look at a diagonal metric, it's obvious that the coordinate speed of light in the x direction is just

$$g_{tt} dt^2 = -g_{xx} dx^2$$

Thus the speed of light will always be c when $|g_{tt}| = |g_{xx}|$ in geometric units.

The speed of light thus depends entirely on the metric used. It is possible for different people to use different coordinates, which give different metrics - and thus different speeds of light.

So, when we ask what the "speed of light" is we do not get an answer until the metric is defined - it's a metric dependent quantity.

A local observer is just an observer who constructs his metric using meter sticks and clocks that he carries with him.

[add]Probably I should add some more detail, but I don't want to spend a lot of time on this unless it's key to what you're interested in. Mentioning that the idea is to set up "Fermi normal coordinates" is probably good enough at this point.

If our local observer using the SI standard meter, he will not even have to measure the speed of light at his location - it will automatically be equal to 'c'. If we go back to the older definition, the local observer would have to cary a physical copy of the Paris meter-stick with him, as well as a cesium clock. The speed of light then becomes a measurable quantity, but our observer will find that at his location, it is always equal to 'c' because the laws of physics are not dependent on velocity, and the effect of acceleration can be made negligible by chosing to measure the speed over a short enough distance.

Last edited: Mar 15, 2006