# Measurements of time/distance be dependent on an object's velocity

1. Jul 3, 2005

### jason_one

I'm just starting to learn about relativty, and I'm having trouble understanding it. Maybe someone could help? The first thing that I'm have trouble grasping is why would measurements of time/distance be dependent on an object's velocity relative to the speed of light?

Last edited by a moderator: Jan 7, 2014
2. Jul 3, 2005

### εllipse

The special theory of relativity has two postulates (assumptions):

1. the principle of relativity (in the restricted sense) - This states that all inertial (non-accelerating) observers are equivalent for the formulation of the laws of nature. (This means that anyone who is moving with a constant velocity can rightfully call himself at "rest".)

2. the constancy of the propagation of light in vacuo - This states that light travels at the same speed for all observers (who are at rest).

So imagine you and me are both floating out in black, empty space far removed from any stars. I call myself "sitting still" and carry out an experiment to measure the speed of light by telling you to go way off in the distance and then stop and sit still. We both have light bulbs (which are turned off), and your instructions are to turn on your light bulb immediately when you see mine come on. So, when I turn on my light bulb, I just count the seconds to see how long it takes me to see your light bulb and then I know how long it took the light to reach you by dividing that number of seconds by 2 (because the light had to travel twice the distance between us). Then by dividing the distance you are from me by the amount of time it took for my light to reach you, I now know the speed of light, and I call this C.

I then tell you to move away from me at 50% of C. Newton, who didn't base his theories on the above postulates, would have said that if I turned on a light bulb, you would claim that the light from it would travel at 50% of C toward you, because every time it moves forward some you also move forward. However, Einstein (based on both Maxwell's equations (which Einstein admired) and early experimental data) knew that the speed of light was constant for all observers. Unlike Newton, Einstein knew that you would not claim that the light only travelled at 50% of C toward you, but that you would still make the claim that the light travelled at the full 100% of C toward you.

This may seem like an unresolvable conflict, but Einstein realized that the way to resolve it was to throw out the unproven idea that all observers measure time and distance the same. Although it seems to us that time should tick along the same for everybody, there is absolutely no experimental data to back this up. Instead, the experimental evidence shows that time does slow down for fast speeds.

Another way to think about it is to imagine chasing a light ray in a very powerful rocket; you slam on the accelerator to try to catch up with the light ray, but are surprised to find that no matter how hard you push on the accelerator you can't gain on the light ray. Not only can you not gain on it, but everytime you measure how fast away it's moving from you, it's always moving at the same speed, as if it speeds up just as much as you do. Of course, the light ray isn't really speeding up, so why can't you ever seem to catch it? Again, the only way to resolve the conflict is to throw out the notion that time and distance are the same for everyone. Time and space must change the faster you go to allow the speed of light to remain constant.

Last edited: Jul 3, 2005
3. Jul 4, 2005

### jason_one

Okay, thanks ellipse, that makes sense, I understand now. I still have another question, though.

So if I, with my lightbulb, am travelling at 50% of C away from you, Newtonian physics would say that C would appear to be 50% slower, but Einstein tells us that C would still appear to be at 100% and that time would be moving more slowly as we approach C.

But what if the opposite scenario happens, where we are still the same initial distance apart, but instead of moving at .50C away from you, I move at .50C towards you. Newtonian physics would say that C would appear to be 150% of what it would be at rest, so to compensate with relativity, wouldn't we say that time for me is speeding up? And then once I passed you, and the light was traveling in the same direction as me, time would slow down again. This would mean that time would either speed up or slow down for me, not dependent on my velocity (because it was .50C in both cases), but dependent on my direction of motion relative to the source of light?

And another question I have is this: Let's say I was traveling near the speed of light in a long (2,000m), dark box in space and I am standing at the midpoint of this box, so that the diagram looks like this:

|________________.________________|
1000m Me 1000m

-----------------------------------------------> v is near c

If I, standing at this midpoint, activate a switch that simultaneously turns on two light bulbs placed at opposing ends of the box, will the light from each bulb hit me at different times? The light from the bulb on the right would have to hit me first because the light has less distance to travel, seeing as I am advancing rapidly toward the point in space where it first emitted its rays. And the longer the box, the more apparent the delay would be. Also, the closer I get to c, the more apparent the delay would be.

But I am not accelarating, I am at a constance velocity, so by the principle of relativity (in the restricted sense) I could call myself at rest. How then would I explain to myself the time delay between each light hitting me?

4. Jul 4, 2005

### Aer

This is a disingenuous statement often used by people knowledgable in special relativity. Maybe the implication was unintention on your part, but nevertheless, your statement seems to imply that time slows down for the object in motion according to the said object in motion. This is what leads to much confusion for people trying to learn special relativity because you'll go on to state that the said object in motion can consider itself at rest and therefore time slows down for the other frame.

5. Jul 4, 2005

### Aer

Time for the object in motion remains the same. Case and point of what I said above.

6. Jul 4, 2005

### Päällikkö

The statement "I am moving" is meaningless.
From your point of view, you are at rest in the dark box. So, you will see the light beams at the same instant.

The other observer, relative to whom you're moving at 0,5c, sees the situation as you described it: One light beam hits you before the other.

7. Jul 4, 2005

### Aer

There is nothing wrong with myself considering the earth at rest while I am moving with resect to the earth. Your argument suggest since I exist in one frame, I cannot assume any other frame is at rest, which is ridiculous. Maybe that is not what you meant and you are considering some other statements are true which implies you must consider yourself at rest. But this is not what you said and it is perfectly appropriate for jason_one to consider rest to be whatever he wants regardless of how "he" fits into this rest frame.

8. Jul 4, 2005

### jason_one

So, Paallikko, let's say my box is at rest in space. I turn the lights on and I see the rays hit me at the same time. Then I accelerate to a speed that is near C. Once I reach that speed and I hit the lights again, how could they possibly hit me at the same time?

If C is constant, regardles of the inertia of my box, the light from the bulb that I am traveling towards should hit me first (because it has less distance to cover before it reaches me since I am traveling toward the point where it was emitted) and the light from the bulb I am traveling away from should hit me last (because it has more distance to cover before reaching me). If they both hit me at the same time, that would mean that the light from the point I am traveling toward is propogating more slowly than the light from the point I am traveling away from, which would be impossible.

The only other way for both rays of light to hit me at the same time then, would be if time was flowing at different rates in opposite halves of the box, which would also be impossible since both are traveling with the same velocity.

This means that the light has to hit me at different times, but since I am at rest (according to principle of relativity), this should also be impossible.

How do I, in my box, understand this dilemma?

9. Jul 4, 2005

### Päällikkö

I see yoru point, but: There is no "ether", relative to which light moves at the speed c. Instead, the speed of light is constant in all inertial frames.

After accelerating, you're again in an inertial frame ("at rest"), thus you will measure the speed of light to be c.

If this wasn't the case, the laws of physics wouldn't be same in all inertial frames -> the foundation of Einstein's theory of relativity would be incorrect.

I am unsure what will happen if you do the test in an accelerated frame, though.

10. Jul 4, 2005

### Janus

Staff Emeritus
Okay, as long as the lights turn on simultaneously according to you, the light from each bulb will reach you at the same time. Light always travels at c with respect to you.
But let's consider what happens according to some one traveling with respect to you. Imagine that he passes you the instant the light from the two bulbs reach you. He will also see the light form the two bulbs at the same time. But for him, the distance to the bulbs does not remain constant. Since he also measures light as traveling at c with respect to himself, he has to determine that the two bulbs did not turn on at the same time. This is the third effect of Relativity: The "Failure of Simultaneity" or "Relativity of Simultaneity". Simply put, events that are separated by distance and are simultaneous in one frame are not always simultaneous in a frame that is moving relative to the first.

11. Jul 4, 2005

### Janus

Staff Emeritus
c is a constant as measured relative to any frame as measured from that frame. Thus in the first instance you will measure light as moving at c "relative to you". In the second instance, you will again measure light as moving at c "relative to you".

12. Jul 4, 2005

### JesseM

No, the other person will still see your clocks slowed down. But the fact that you will always measure the speed of light to be c is not explained solely in terms of the fact that your clocks slow down (from the other person's point of view), it's a combination of three things:

Time dilation: the other person sees the ticks of your clocks lengthened by a factor of $$1 / \sqrt{1 - v^2/c^2}$$

Lorentz contraction: the other person sees your rulers shrunk down by a factor of $$\sqrt{1 - v^2/c^2}$$

The relativity of simultaneity: if you have a bunch of clocks in a row which are all "synchronized" in your own rest frame, then in the other person's frame the clocks are all out-of-sync, such that for any two clocks of yours that the other person sees as being a distance x apart, the other person will see the trailing clock ahead of the leading clock (note that this depends on the direction of your velocity) by $$\gamma \frac{vx}{c^2}$$, where $$\gamma = 1/\sqrt{1 - v^2/c^}$$

Note that if the other person sees the clocks as being a distance x apart, you see them being a larger distance $$x' = x \gamma$$ apart, so we could also say that for any two clocks that you see a distance x' apart, the other person will see the trailing clock ahead of the leading clock by $$\frac{vx'}{c^2}$$

When you measure the "velocity" of a light ray, it's assumed that you do this by having a bunch of clocks in a row along a ruler that's at rest relative to you, with the clocks being "synchronized" in your frame (The synchronization procedure is to set off a flash at the midpoint of two clocks, then set them so they both read the same time at the moment the light from the flash reaches them--this synchronization procedure will result in the clocks being out-of-sync from the perspective of an observer moving relative to you, since he'll assume light travels at c in both directions in his frame, and since he sees one clock moving toward the flash and one clock moving away from it, he won't think the light hits both at the same moment.) So, to measure any object's velocity, you look at the reading of the clock sitting at point A on the ruler at the moment the object passed point A, then you look at the reading of the clock sitting at point B on the ruler at the moment the object passed point B, then you take the distance between A and B as measured by the ruler and divide by the difference between the two clock's readings. So, from the point of view of someone moving with respect to you, the result you get for this measurement depends on how much your clocks are slowed down, how shrunken your ruler is, and how far out-of-sync clocks at different points on your ruler are.

Here's a simple example. Suppose, for the sake of making the math a bit easier, that we measure distance in units of "fivers", where a fiver is defined to be the distance light travels in 0.2 seconds, so that light is defined to have a velocity of 5 fivers/second. Suppose you see a ruler which is moving right-to-left at a velocity of 3 fivers/second towards you along your x-axis. In its own rest frame, this ruler is 40 fivers long; so in your frame its length will appear to be: 40*squareroot(1 - v^2/c^2) = 40*squareroot(1 - 9/25) = 40*0.8 = 32 fivers long. Also, at either end of this ruler is placed a clock; using the time dilation formula, we can see that for every second on your clock, you will only see these clocks ticking 0.8 seconds forward.

Now, say that when t=0 according to your clock, the clock on the right end of the ruler also reads t'=0, and is at positon x=100. Since the ruler is 32 fivers long in your frame, this means the left end will start out at x=68. At that moment, a light is flashed on at the right end of the ruler, and you observe how long the light pulse takes to catch up with the left end. In your frame, the position of the light pulse along the x-axis at time t will be (100 - c*t), while the position of the left end of the ruler at time t will be (68 - v*t). So, the light will catch up to the left end when (100 - c*t) = (68 - v*t) which if you solve for t means t = 32/(c - v). Plugging in c = 5 and v = 3, you get a time of 16 seconds for the light to catch up with the left end, in your frame.

Now, using the formula for the relativity of simultaneity, we can see that if two clocks on the moving ruler are 40 fivers apart in the ruler's own rest frame, then in your frame the clock on the left is always (3)*(40)/(25) = 4.8 seconds behind the clock on the right, so when the clock on the right reads t'=0 the clock on the left reads t'=-4.8. And the time dilation formula tells us that after 16 seconds have passed according to your clocks, only 16*0.8 = 12.8 seconds will have elapsed on each of the ruler's clocks, which means the clock on the left reads -4.8 + 12.8 = 8 seconds at the moment that the light reaches the left end. So remembering that light was emitted when the clock on the right read t'=0, the light must have taken 8 seconds to cross the ruler in the ruler's own frame; and remembering that the ruler is 40 fivers long in its own frame, the speed of the light pulse is measured to be 40/8 = 5 fivers/second. So, light does indeed have the same speed in both frames.

Last edited: Jul 4, 2005
13. Jul 4, 2005

### jason_one

Hmm... I'm still having trouble understanding this.

Let's say the box is 372,000 miles long (twice the distance light propages in one second), and instead of a bulb on both ends that I can activate, I have two men with indentical clocks who are instructed to both turn on a light when the clock strikes 12. I also have the same clock and I have to measure the amount of time it takes for both beams to reach me.

When we are at rest, they turn the lights on at 12:00 and I record the light hitting me one second after that.

But if we are in motion, something close to C, what will happen?

The solid line represents the box moving at constant velocity (v); the M represents me and my clock at the midpoint; the verticle lines represent my men and their clocks at the end points; the asterisks (*) represent the points in space where the two flashes were emitted; and the lines with arrows represent the propagation of light from those two points.

|____________M_____________|----->(v) 12:00
|*..................M...................*|

......|____________M_____________|----->(v)
|*---->........................<----*

..........|____________M_____________|----->(v)
|*-------->.............<--------*

...............|____________M_____________|----->(v) One second later
|*------------>..<------------*

If my men are both instructed to emit a flash when our identical clocks strike 12, wouldn't I record the flash on the right has happening first? In order for me to record both flashes hitting me simultaneously after one second and for both events to be simultaneous for my men as well (since we are all traveling at the same speed), the diagram would look like this:

|____________M_____________|----->(v) 12:00
|*..................M...................*|

......|____________M_____________|----->(v)
|*------->........................<-*

..........|____________M_____________|----->(v)
|*-------------->.............<--*

...............|____________M_____________|----->(v) One second later
|*---------------------->..<--*

In order for the propagation (distance\time) of light to remain constant (and for these simultaneous events to appear simultaneous to all observes in the box), this would have to mean that time both sped up and slowed down in different parts of the box, wouldn't it?

14. Jul 4, 2005

### Staff: Mentor

15. Jul 4, 2005

### Staff: Mentor

That is, the two light pulses are emitted simultaneously, and they arrive at you simultaneously.

Since you said "we are in motion", I assume that your assistants at the ends of the box are moving along with you. In that case, in your (and your assistants') reference frame, everything behaves just like before. The light pulses leave the ends of the box simultaneously, travel at the invariant speed c in your reference frame, and they arrive at your location simultaneously.

There is no way that you can conclude that you (and your assistants) are in motion, without "looking outside" your reference frame. Think of riding on an airplane. If the flight path is perfectly straight and level, and the plane is flying at constant speed, any physics experiment that you can do inside the airplane's cabin, that does not make reference to what's happening outside (e.g. by looking out the windows), gives the same result as on the ground.

In the reference frame of a stationary observer on the ground, things are somewhat different. In this reference frame, the light at the rear end of the box comes on before the light at the front end. Both pulses travel at the invariant speed, c, but the light from the rear end has to travel further to reach you than the light from the front end, because you're moving away from the point at which the rear pulse was emitted, and towards the point at which the front pulse was emitted. Therefore the rear pulse has to travel for a longer period of time to reach you, than the front pulse does. This exactly compensates for the difference in emission times, and the pulses arrive at your location simultaneously in this reference frame also.

If your assistants turned on the lights at 12:00, according to clocks that you (and they) carry, and are synchronized in your (moving) reference frame, then those clocks are not synchronized in the ground observer's reference frame. In his frame, both (moving) clocks run at the same rate, but more slowly than his own clock, and the rear clock leads the front clock (that is, it shows 12:00 first).

16. Jul 4, 2005

### Päällikkö

No.
You are not moving relative to the black box you're in.
The case'd be different if you were outside the box and the box were moving relative to you. As bizarre as it may seem, light travels at the speed c in all inertial frames.

Earth moves around the sun, shouldn't eg. microwaves act silly when it's moving to different directions? And shouldn't we on earth measure the speed of light different from different directions? NO, states Einstein's principle of relativity.

There are whole galaxies moving some 0,8c away from us. It makes no sense to think microwave ovens wouldn't work there.

PS. There is NO absolute motion. You cannot say: "I'm moving at 0,8c". However, you can say: "I'm moving 0,8c relative to Earth". You cannot gain on light. no matter how fast you go (relative to eg. Earth), you will always see light moving at the speed c relative to you (assuming you're in uniform motion).

PPS. Think about tennis. If you play tennis on a ship (preferably under deck, where there's no wind), do you have to take the direction and speed of the ship into account? No, this is the principle of inertia.

Last edited: Jul 4, 2005
17. Jul 4, 2005

### Janus

Staff Emeritus

Maybe this will help:
http://www.geocities.com/janus58.geo/simultaneous.html

18. Jul 4, 2005

### Aer

Yes.
What is in motion? And what is it in motion with respect to?

NO. This is wrong in every sense of the word. You seem to want to attribute an absolute rest point to the v=0. This is unfounded - there is no such thing as absolute rest.

19. Jul 5, 2005

### JesseM

Well, there's no such thing as absolute velocity in relativity, there's always going to be some frames where you're moving close to c and another frame where you're at rest. But we can ask what will be seen by an observer in a frame where the box is moving close to c (of course, in the box's rest frame, it's the box that's at rest and the observer who's moving at close to c). This observer will see the two clocks being out-of-sync, so the clock at the back of the box ticks 12:00 before the clock at the front of the box ticks 12:00, in his frame. Because of this, he will also see both light beams hit the person at the center at the same moment, despite the fact that the person at the center was moving away from the flash at the back end of the box and towards the flash at the front end. If you want an actual numerical example, just specify how fast the box is moving in this outside observer's frame.
If the box is moving to the right in the outside observer's frame, then this observer will say the flash on the left end happened first--after all, since the guy at the center is moving away from the flash at the left end and towards the flash at the right end, then if light is assumed to travel at a constant velocity in this outside observer's frame, the light from the flash on the left will take longer to catch up with the guy in the center than the flash on the right, so the only way they'll hit him at the same moment is if the flash on the left happened earlier.
Nope, all that's required is that the clock on the left end hit 12:00 before the clock on the right, but this is because they are out-of-sync by a constant amount, not because they are ticking at different speeds. The outside observer will see both clocks ticking slow, but the one at the back will always be ahead of the one at the front by $$vx'/c^2$$, where x' is the distance between the clocks in the box's rest frame, and v is the velocity of the box in the outside observer's frame.

20. Jul 6, 2005

### EnumaElish

A question

This makes sense as a thought experiment; but I thought (used to think) that time dilation was not directional. I.e. two clocks aboard the box would be in sync in all frames because both are moving at the same speed. Do you think that may be due to confusing speed and velocity?