# Comparative speed of moving vs unmoving

1. Oct 11, 2011

### gonegahgah

I've done a little diagram to hopefully represent my question.

I've represented three different sources on three moving vehicles:
Source 1: cannon (emitting cannon ball)
Source 2: horn (emitting sound)
Source 3: light (emitting light)

For the cannon the cannon ball will reach the target sooner than a cannon ball fired from a stationary cannon.
For the horn the sound will reach the target at the same time as sound from a stationary horn.
For the torch the light will reach the target (at the same time)/(sooner than) light from a stationary torch?

Could someone with enough credentials just clear up for me once and for all what the answer is for the last one? Will the light from the moving torch reach the target at the same time as the light from a stationary torch or would it reach the target sooner than the light from the stationary torch?

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2. Oct 11, 2011

### ghwellsjr

Same time.

3. Oct 11, 2011

### gonegahgah

Cool, thx. What about if the target is moving vs not moving instead of the source - say towards the emitter?

For the cannon ball the situation remains the same.
For the horn the sound will reach the target moving towards it sooner than the target that stays still. (This is different and is due to the non-movement of the air).
For the torch the light will reach the target moving towards it (sooner)/(at the same time) as for the target that stays at the line?

Which of these is the correct one for the torch and light in this example?

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4. Oct 11, 2011

### ghwellsjr

Two light beams, starting at the same place but without regard to the relative motion of whatever is emitting them, will travel through space together and arrive at any location coincidentally.

5. Oct 11, 2011

### gonegahgah

Thx. I'll try to understand what you are saying then?

For the second example the moving target will be at a different position than the stay put target when the light reaches the closer one so that would mean the light will reach the moving target sooner?

Is that what you mean?

6. Oct 11, 2011

### ghwellsjr

The motion of the sources and targets have nothing to do with the propagation of light. But the light is traveling in a straight line so any targets, moving or not, that are closer to the source of light, moving or not, will receive the light before targets that are farther away.

7. Oct 11, 2011

### gonegahgah

Thx. The reason I am asking is I am trying to work out if I have looking at it all wrong all this time and if maybe things can fall into place in a way that I can be comfortable in accepting. I'm not totally sure of this but I'm hoping.

But I'm thinking about it and I'm not sure now...
I thought maybe it had something to do with the observer - which is what is stated but it doesn't fix things the way I was hoping...

I have trouble seeing any difference between the source moving towards the target or the target moving towards the source yet they produce different transmission time results for the same initial separations.

The cannon ball example is easy and is just added velocities.
The sound example is easy and relates purely to the movement of air and the speed of sound in air. If you produce a sound in a wind it will carry faster in the direction of the wind and slower against the direction of the wind. Speed of sound is relative to the body of air.

But, light speed through vacuum is not affected by any medium or the movement of that medium; only the observer; yet the results more closely parallel the sound example and not the cannon ball example? Why? I'm confused.

If you move towards the Sun then you expect to meet younger light (time elapsed since it was emitted) then a ship that stays behind. But if you have two Sun's, one moving towards you and one staying put, then if they pass through the same point light from both will reach you at exactly the same time.

Or if we drop it to a single ship and sun example: If you move towards the Sun then you expect to meet younger light than if you had stayed put. But if instead the Sun moves towards you or not then the 'age' (time since emission) will be the same.

To me these two situations sound exactly the same but for whatever reason:
- if you are moving towards the Sun you will reach 'younger' light
- if the Sun is moving towards you then you will receive light that is the same 'age' as light from a Sun that weren't moving...

It sure has me confused??? If I'm mixing something up badly can you help me?

8. Oct 11, 2011

### Staff: Mentor

Different transmission times according to whom? Who is measuring the transmission time?

(And realize that this is different than asking which light hits the target first.)

9. Oct 11, 2011

### ghwellsjr

The movement of light through space is not affected by an observer. We don't know how light moves through space, only that two parallel light beams or photons travel together. We have no way of tracking the progress of light. We don't know when the light reaches a given location. We don't know where the light is at any given moment in time. And the reason is that we have nothing that can travel faster than light to give us information about its whereabouts like we do for cannon balls or sound where we use lightwaves to track their progress.

So to solve this otherwise intractable problem, Einstein came up with the novel idea of rearranging our concepts of time and space so that we instead define a Frame of Reference in which the speed at which light propagates through space is c. It's a Frame of Reference in which the movement of light is now knowable, whether or not there are any observers at rest in that FoR.

10. Oct 11, 2011

### gonegahgah

You're right Doc. This is the part I'm trying to think into it; how the observer can make the difference.

I'm thinking that the observer is outside the emitter and the target rather than being in either. So,

- If the outside observer is moving in step with the emitter (so that the emitter appears stationary) then the result is supposed to be the same amount of time that the light reaches the target (whether the emitter moves or not),
- or if the outside observer is moving in step with the target then the result is supposed to be 'younger' light for the target that is moving towards the emitter as opposed to if the target doesn't move.
(*I've just reread these and I may have them back to front. I'll have a rethink about them tomorrow. Maybe there is something for me understand here? Not sure yet?)

But why? I was hoping the observer's movement somehow affected the result mathematically but I'm failing to register how?

Using the observer as a mathematical foundation I would have expected similar results but in the opposite way than they appear. So I'm still confused. I'm a bit sleepy but I'll try to understand it a bit better tomorrow...
(*As mentioned I'll recheck my thinking tomorrow; maybe there is hope?)

The tricky bit may be (?) that we use light to measure light which certainly pertains to the observer and maybe that is what is confusing me... Thanks for assisting me; I'll look again tomorrow.

11. Oct 11, 2011

### ghwellsjr

As I said before, we use light to measure cannon balls and sound waves but we cannot use light to measure light, we need something faster if we are going to measure the propagation of light in space but there is nothing faster and that's how Einstein comes to our rescue, by redefining what time and space are so that we can talk meaningfully about the propagation of light through space.

12. Oct 11, 2011

### HallsofIvy

Staff Emeritus
The "cannon ball" and "light" examples are not all that different but you are using the wrong formula.

If a vehicle is moving, relative to the ground, at speed u and you fire a cannonball forward at speed, relative to the vehicle, v, then the cannonball's speed, relative to the ground is NOT simply "u+ v", it is, according to special relativity,
$$\frac{u+ v}{1+ \frac{uv}{c^2}}$$
Of course, for normal "vehicle" and "cannonball" speeds u and v are so small compared with c that $1+ uv/c^2$ is essentially 1.

If you shine light forward, at speed c relative to the truck, by the same formula, its speed relative to the road is still
$$\frac{u+ c}{1+ \frac{uc}{c^2}}= \frac{u+ c}{1+ \frac{u}{c}}= (u+ c)\frac{c}{u+ c}= c$$

13. Oct 11, 2011

### gonegahgah

Thanks Ivy. That's a cool formula. Thanks for showing it for both to clarify the similarity.

I am still confused...
There still seems to be the matter that if you, the observer, are moving with the target you will be at a closer point to the light when the light reaches you travelling at c so the final distance is shorter; whereas if you, the observer, are moving with the emitter then distance to the target remains the original distance and so will take the full amount of time to travel to the target.
For the cannon ball this doesn't matter and the final distance covered is always shorter than the original distance and so the time is shorter whether you, the observer, are moving in step with the emitter or the target.

This is why I'm trying to think how the observer fits into the light example.
If the outside observer moves in step with the target as it closes on the emitter then they will get a report of the light sooner - than if the target and emitters weren't moving - because they will move towards the light as the light travels towards them; so they meet part way.
But if the outside observer moves in step with the emitter then the light travels to the target for the full amount of time dictated by the distance but the report back of the target being hit would be affected by the movement of the observer. The report back to the outside observer now takes less time to get back because the observer is moving towards the target.

For this later case will they measure expiration of the full time for the original distance plus the time for the report to get back to the observer where they are now closer to the target because they are moving in step with the emitter. Using those relative formulas is that correct?

14. Oct 12, 2011

### ghwellsjr

Up until now, you've been asking about the one-way propagation of light but now you're introducing an outside observer without making it clear where in relation to the emitter and the target he is located so it's not possible to provide you with a simple answer. Why don't you stick with the one-way propagation issues until they make perfect sense to you?

For purposes of understanding the one-way propagation of light, you can consider it to be exactly like your analogy with the propagation of sound in air. So let's work out some examples. In these examples, you and I will be the targets listening for the sounds from three horns, each with a different pitch so we can distinguish them.

First, let's take a situation where you are stationary with respect to the air and I am going to be moving but I will explain how later. The first horn is stationary. The second one is moving toward you and the last one is moving away from you but they are all far away from you and in line with each other and with you. It just so happens that all three horns arrive at the same location at the same time and they each emit their respective sounds. Isn't it obvious that all three sounds will travel together through the air and arrive at you simultaneously? Now suppose we repeat the experiment but this time, I'm traveling toward you from farther away than the horns and I happen to arrive at your location at just the moment that you hear the three horns. Isn't it obvious that I also will hear all three horns at the same time too? And if I were approaching you from a position closer to the horns than you are but arrived at your location at the moment you heard the horns then I would hear all three horns at the same time as you did too?

OK, got all that? Good.

Now let's repeat the whole thing but this time the wind is blowing in the direction from the horns toward you. Will that change the order in which the sounds of the horns reach either you or me? No, they will still all arrive simultaneously both for you and me, won't they? Same thing if the wind is blowing in the opposite direction or any other direction. It also won't matter how fast the wind is blowing, will it?

Now, instead of having the wind blow in different ways, let's just have you, me and the horns all moving with respect to each other, thus creating our own wind. This, again, won't make any difference, it terms of the simultaneity of the arrival of the sounds of the horns, will it? (I'm assuming that we don't go so fast as to break the sound barrier.)

This is exactly like our situation with light. When two or more light sources traveling with respect to each other emit flashes of light when they are co-located, those flashes propagate through space together and arrive simultaneously at co-located targets, no matter how the sources or targets are moving.

Please note that we are not comparing the time it takes for the sound to travel from the horns to you and me under different conditions of the wind because that will definitely affect the propagation time and we're not measuring the propagation time. We're only demonstrating that it isn't affected by motions of the emitters or targets. In the case of sound in air, we can know the actual propagation time for the sounds from the emitters to the targets because we can use light to let us know when the horns were tooted. But we can't do that with light for the same reason we couldn't use the sound waves all by themselves to measure the propagation of other sound waves.

15. Oct 12, 2011

### gonegahgah

Thanks that is good. You've no-noed it but I like that idea of using sound to measure the arrival time of sound at the target as this is kind of like what I am wondering for light. The communication of arrival time has to be done by some means and generally that is emr (including current in a wire). But as you say, I should take little steps to see if I can understand this.

What you are saying is making sense; though again light doesn't have any media in a vacuum to affect its speed whereas sound does as you say. I'll give this some more thought tomorrow. Thanks ghwellsjr.

16. Oct 12, 2011

### ghwellsjr

I hope you aren't thinking that we are measuring the arrival time of light. We aren't doing that. All we are doing is illustrating that the propagation of light through space is independent of the motion of the source or the destination. I haven't said anything about measuring anything. And I have specifically said that we cannot measure the propagation of light because we don't have anything faster than light with which to measure it.

17. Oct 12, 2011

### harrylin

When you talk about "the observer's movement", you imply "as seen by a different observer". And indeed, according to that other observer, the motion of the moving observer affects the result mathematically, because the moving measurement system is affected by motion.
However, from the perspective of the "moving" observer who assumes to be in rest, it's just the other way round.

PS: And as mentioned before, SR treats light just the same as sound in air that is in rest.

Last edited: Oct 12, 2011
18. Oct 13, 2011

### gonegahgah

That is sounding like how it works harrylin with the air/space-time being at rest relative to the observer or something like that? I think? I'm guessing?

Another question...

When you head towards a light wave you meet the light wave part way but how much time does it take to reach you. Does it take a major fraction of the time that it would have taken if you hadn't moved from your starting point; or does it take the same amount of time to reach you?

Like the cannon ball. If you move towards the cannon you will meet the cannon ball part way and the time elapsed to being hit is less than if you stayed put instead of moving towards it.

And the sound wave. If you move towards the horn you will meet the sound part way and the time elapsed to hearing the sound is less than if you stayed put instead of moving towards it.

What is the situation for a light wave? If you move towards a torch you meet the light part way but does the light take less time to reach you than if you stayed put? Or does it take the full amount of time of your original separation?

19. Oct 13, 2011

### harrylin

If you still use the standard measurement system in which you were in rest also when you are moving, then light will behave just like sound in air that is in rest in that system.

However, if you immediately after you start to move towards the torch, adapt your measurement system to one in which you are then in rest, you will recalibrate and resynchronize your measurement system in such a way that you'll measure the light to move at c towards you; however the distance that the light has to travel will appear less in that system.

So, with either of these choices you will measure that the light will take less time to reach you.

Last edited: Oct 13, 2011
20. Oct 13, 2011

### ghwellsjr

When you say "measurement system", do you mean anything other than what everyone else means when they say "Frame of Reference" as defined by Einstein's Theory of Special Relativity?