# Basics: Speed

1. Jun 8, 2007

### Little Franklin

Hello.

My understanding of this subject is limited. I thought I basicly understood it, but now it seems my concept of it was similar to the theory of the http://en.wikipedia.org/wiki/Luminiferous_ether" [Broken]. I've read more about it but I still don't understand, and wondered if I could ask some questions?

What does light move at the speed of light relative to?

Last edited by a moderator: May 2, 2017
2. Jun 8, 2007

### Staff: Mentor

Any inertial observer.

3. Jun 8, 2007

### George Jones

Staff Emeritus
Any local observer, where a local observer is an observer, inertial or not, which the light whizzes by.

4. Jun 8, 2007

### Little Franklin

If I've understood what I've read right, mass can slow time and therefore light. So although a photon would travel at the speed of light as it reached an observer, it could also be observed to have taken longer than it should have to reach the observer if it passed a large mass, or shorter if observed from a large mass.

Is this correct?

5. Jun 8, 2007

### Chris Hillman

No wonder you are confused: "mass can slow time" makes no sense. (If time itself somehow ran slower, what clock would know?)

You are no doubt referring to what is sometimes called the "gravitational red shift effect" in gtr, a theory in which we model spacetime as a curved Lorentzian manifold (unlike str, where we ignore gravitation and model it as a flat Lorentzian manifold). In a curved Lorentzian manifold, the world lines of light signals (traveling through a vacuum) correspond to null geodesics. One effect of curvature is that two initially parallel geodesics (these are the analogues of "straight lines" in flat spacetime) can spread apart. The result is that time signals from an observer close to a massive object appear red shifted when received by an observer who is further away.

When you said "it could also be observed to have taken longer than it should have to reach the observer if it passed a large mass", you appear to be referring to a different but related effect, the Shapiro time delay effect. This again involves the behavior of geodesics in the large. We very recently discussed this in another thread; use the PF search tool to look for posts mentioning "Shapiro light delay". As I emphasized there and elsewhere, many phenomena in gtr involve notions of "distance in the large", and thus cannot be properly understood without being aware that there are multiple distinct but operationally significant notions of distance in the large, and thus of "speed in the large". However, "speed in the small" always works just the same way in gtr and str. That is why it makes sense to say "the speed of light in vacuo" is c in both gtr and str.

Last edited: Jun 8, 2007
6. Jun 8, 2007

### pervect

Staff Emeritus
It's probably correct, but let's add some more detail into the thought experiment to make the example more complete.

Suppose we look at the Shapiro effect time delay, where we bounce a radar signal off of Venus from the Earth.

When the radar signal passes very close to the sun on its way to and from venus, GR predicts that that the signal propagation delay, as measured from Earth, increases. This effect is measured by using clocks located on the Earth that measure the total trip time of the radar beam, much as any radar does. There are a couple of different ways of doing this, but I don't think we need to get into the engineering.

If we hypothetically placed observers all along the radar beam all of them would, however, measure that the speed of light was equal to 'c' using their own clocks and rulers.

The usual way of explaining this is to say that the clocks tick at different rates when they are near or far away from a large mass. An alternate way of looking at things is to say that the clocks all tick at 1 second per second, but that the comparison process makes them appear to tick at different rates. If you're fairly new to relativity and don't quite see why anyone would prefer the second explanation I'm not surprised, but it does have some advantages.

7. Jun 8, 2007

### MeJennifer

There are alternative ways to look at it:

Quantum mechanics shows us that all matter has a frequency proportional to the Planck constant. One could consider any displacement between matter a gravitational potential energy which is compensated for by a change in this frequency (e.g. the relative speed of clocks).

Richard Feynman formulated it this way:
“Gravity is that field which corresponds to a gauge invariance with respect to displacement transformations.”

Last edited: Jun 8, 2007
8. Jun 9, 2007

### RandallB

Good point even an accelerating observer will measure c the same.

9. Jun 12, 2007

### Little Franklin

I don't understand that as I will now demonstrate.

You say the frequency compensates for the potential energy which I assume means they're inversley proportional. But doesn't something closer to a source of gravity have less potential energy, meaning a higher frequency which would mean higher speeds.

Since I don't even understand GTR I should probably stay away from Quantum mechanics for now.

They seem to be the same thing to me which I surpose means I haven't understood the second explanation.

What are the advantages of the second one?

10. Jun 12, 2007

### MeJennifer

The gravitational potential energy is negative.

Consider a region with some mass-energy, split this region and create a spatial separation between them. The mass-energy in both regions increases due to the increased gravitational potential energy between them.

Interestingly enough, consider clock A and B on the Earth. Now bring A to the second floor of a building, then we know that A is ticking a very small amount faster than before. But also B must now be ticking an even smaller amount faster than before since A's mass was spatially separated from Earth.

Last edited: Jun 12, 2007
11. Jun 12, 2007

### George Jones

Staff Emeritus
Suppose that you and I go to a store together and buy buy identical watches, and that some time after this we part ways. I (use a rocket to) hover above a large mass and you hover above (i.e., I am between you and the mass) me. With one eye, you look through a telescope and watch the second hand on my watch, and you watch the second hand on your watch with your other eye. The image that you see of my second had spins more slowy than the image you see of your second hand.

Sometimes, people summarize this very loosely by saying "George's watch runs slower than Little Franklin's watch." I prefer not to say this. Our watches are identical, and we each think that our watch is running at its normal rate, when compared to stuff like how fast our heart beats and how fast our hair turns grey. When we were together at their purcahse we saw the watches running at the same rate, and neither of us noticed anything funny happen to the rate of running of our own watches after we separated.

I spite of all this, as I said above, when you use your eyes to compare the images of our watches, the two images that you see whirl at different rates

12. Jun 15, 2007

### Little Franklin

There's the old idea about the two identical twins. One goes off in a rocket at near the speed of light, when he returns the other twin is older than he is.

My question is what makes the rocket the place where time passes slower? How can you say it's moving faster than the planet? Aren't they just two objects moving towards or away from each other?

13. Jun 15, 2007

### rbj

this is what we call the "Twin Paradox". the difference between the two twins is that one is accelerated and the other is not. if the first twin got in a rocket and flew to Alpha Centauri (and stopped there) and later the second twin got in a rocket and flew to the same place, they would view each other as having the same age. but if the second twin never flew to anywhere and the first twin flew back from Alpha Centuri to Earth, then, when they meet, that space-traveling twin would be younger than his/her earth-bound sibling.

14. Jun 18, 2007

### Little Franklin

Does that mean it's acceleration that causes the difference and not speed?

15. Jun 18, 2007

### RandallB

No not at all, the only important part of the twins issue is that one changes or uses two different reference frames. With one remaining in a reference frame, the only way the two can meet to compare watches or aging differences is if the other uses more than one frame to maneuver the back. The accelerations are not important; the time spent in each different reference frame is what counts. Which is why instant speed changes with impossible accelerations work so well when working those problems – you will wind up with fewer significant reference frames to calculate.

16. Jun 20, 2007

### Little Franklin

I surpose what I'm trying to understand is what the difference between reference frames is. You mention speed, but I thought speed could only be measured as being relevent to something else. So my question in this example is what is the speed of the twin in the spaceship relevent to? If the answer were, the twin who stays on a planet, that would seem to imply that the planet was stationary.

17. Jun 20, 2007

### RandallB

Of coarse both twins are stationary and as long as they remain so you can one direct simultaneous side by side look at both of them (when the start at t=0). If they are ever to compare differences directly (meet each other) again, somebody must not remain “stationary” in just one frame. That speed / direction change means a third reference frame for somebody. The one changing reference frames has a new experience relative the other twin and relative to the other reference frame that was “stationary” prior to the speed change.

18. Jun 20, 2007

### petm1

What is the difference between reference frames?

19. Jun 22, 2007

### Little Franklin

Clocks moving at different velocities count time at different speeds relative to each other. So presumably you could find a velocity at which a clock counted as fast as possible relative to the others. What would this velocity represent?

20. Jun 22, 2007

### pervect

Staff Emeritus
No, you can't.

I have the impression that you are falling into the usual trap of disregarding the realtivity of simultaneity.

In order to compare two clocks at all, one needs to define some means by which the comparison can be made, a way to compare the time at which different events occur.

One of the tricky points of relativity is that there is no universal way to compare two distant clocks.

So it is perfectly possible (and in fact, it is the case) for A to think that B's clock runs slow, and B to think A's clock runs slow. A and B assign different meanings to "at the same time", if A marks a set of points "at the same time", he will not have the same set of points that B would mark.

Because the notion of "at the same time" depends on the observer, the process of comparing times also depends on the observer.

You seem to be still thinking in terms of some sort of "universal time" - there is no such thing in relativity, which is why you are getting confused.

So when you think about A's clock running faster than B's, or vica-versa, you need to step back a bit. How do you know that one is faster than another? Learn about space-time diagrams, and draw them.

21. Jun 27, 2007

### Little Franklin

Thanks.
What circumstances would this happen under?

Where is a good place to learn about space-time diagrams?

With the clock experiment I sugested, I understand light emited by an object in a greater space-time curve (not sure what the correct decription is) should be red shifted. Couldn't you then measure the relative passage of time by comparing the red/blue shift of the light from the different objects?

One quick off topic question, is the path of light of higher wavelengths curved more by gravity than that of lower wavelengths, and if so is the effect the same as that of refraction through water/glass?

EDIT: Sorry I only just read about frame dragging. I'll have to think about it some more but I think it may have answered alot of my questions.

Last edited: Jun 27, 2007
22. Jun 27, 2007

### pervect

Staff Emeritus
A space-time diagram is just a graph of position vs time. Usually, though, one makes the time axis vertical in a space-time diagram. This doesn't actually matter, though, it's just the way the graph is usually drawn.

You'll see several examples of space-time diagrams in the FAQ, and in the wikipedia entries.

http://en.wikipedia.org/w/index.php?title=Minkowski_diagram&oldid=136675389

the relativity of simultaneity entry in the wikipedia is also useful and has some space-time diagrams

http://en.wikipedia.org/w/index.php?title=Relativity_of_simultaneity&oldid=138735828
I'm not sure what you mean, but I'll describe one way to measure time dilation, which I think is what you're ultimately after.

What you do is this. You, as the lab observer, want to measure time dilation on a moving clock. What you do is set up a huge array of clocks, one every mile. You then carefully synchronize all these clocks. This synchronization operation is frame dependent - i.e. the moving observer will have a different notion of how to synchronize the clocks.

Now, what you do to measure time dilation is that whenever the moving observer passes over the exact same position as one of your huge array of clocks, you compare them on his moving clock with one of your lab clocks while they are exactly at the same point in space. This is how one can measure "when" a distant event occurs. One uses a clock at the location of the distant event, and synchronizes it to some master clock.

The synchronization process is a key element to make this procedure work. Did I mention it was frame dependent? (I think I did, but I want to stress that point!).

Doing this, you will find that the clock on the moving observer always reads lower than his lab clock does.

This is time dilation.

The moving observer can perform an identical experiment and will get the same result - he will find that what we have been calling the "lab clock" always runs slow.

The key to reconcilling these results is the fact that notion of synchronizing the clocks depends on the observer, i.e. the lab-frame clock synchronization is different than the moving frame synchronization.

Nope. There is no frequency dependence in gravitational lensing.

Last edited: Jun 27, 2007
23. Jun 28, 2007

### MeJennifer

Would you mind telling us where it is shown that a high and a low frequency photon contributes identically to spacetime curvature?

24. Jun 28, 2007

### pervect

Staff Emeritus
As soon as you show me where I said that it did?