# Some Musings

1. Aug 24, 2004

If we take a photon as our frame of reference is it safe to say that the universe moves relative to "Us" at the speed of light?

that is, something 300,000 kms away from us in the "direction" we are "travelling" in, will we reach it in 1 second, or rather since we're stationary, will IT reach US in 1 second?

so what of an object simply travelling towards us at 300,000 km/s (stationary) and an object travelling 450,000 km/s towards us (moving to intercept us at 1/2 light speed). in order to measure light speed as constant for both, we need to contract our distances or dialate our times in these cases. but why are we doing so in the first place?

i realize that relativity theorists will probably retort with some useless saying such as "you can't take the reference of the photon as your frame" or "time doesn't exist for photons, so space-time has no meaning" or something to that extent, but it would be nice if anyone had any interesting thoughts on the matter.

2. Aug 24, 2004

### pervect

Staff Emeritus
Nope, this isn't safe at all. A photon doesn't have a valid reference frame.

3. Aug 24, 2004

### ArmoSkater87

Just like you said in your post, you cant have a reference frame of a photon and neither time nor space exist. Special Relativity doesnt predict what happens at the speed of light, it can only predict what happens as you approach it. From your example, you have 2 speeds, one 300,000 and another 450,000 km/s. Neither can happen, one is AT c, and the other is above c. So, this example cant be used to say things about time dilation and length contraction. Also, you cant add velocities the way that added the two speed to get 1/2 the speed of light. Not only because its impossible to be at those speeds, but because of relativistic velocity addition...
$$u' = \frac{u-v}{\sqrt{1-\frac{uv}{c^2}}}$$
That is the equation for velocity addition of velocity u and v, moving in the same direction, and u' is the relative velocity of one in respect to the other. To calculate the relative velocity of two things going in opposite directions, then change both minus signs to pluses.
If you really wish to know about the reference of a photon, I'll tell you, but again...SR is not involved with it in any way. From the reference frame of a photon, other photons are traveling at infinite speeds.

4. Aug 25, 2004

is that because time = 0?

another thing about the way "speed" is measured to begin with. you say speeds above 300,000 km/s are impossible, but what is that in relation to? If one space ship is going left at 200,000km/s and the other is going right at 200,000 km/s then relative to each other they are seperating at 400,000 km every second that passes.

seems that 300,000 was arrived upon as an arbitrary number (yes i know about Maxwell's) that everyone decided to make formulas to conform to.

they don't really have any basis in logic

5. Aug 25, 2004

### ArmoSkater87

I dont really know the reason.

Wrong, relative to the other, it would not be 400,000 km/s. NOTHING can go the speed of light relative to ANYTHING else. Did you not see Einstein's velocity addition equation that I gave you??? If you use that equation to measure the relative velocity, then you will see that they dont add up the same way that small everyday velocities add up. Relativistic velocities have completely different properties. Use the equation to find the relative velocity of two things moving at .9999999c away or toward each other, and you will see that its not greater than or equal to 1c.

Last edited: Aug 25, 2004
6. Aug 25, 2004

### Chronos

Hmm, logic is not the first word that popped into mind. Bad things happen anytime you let the kids play with zero.

Last edited: Aug 25, 2004
7. Aug 25, 2004

well explain to me this then.

the earth is one reference frame. from the earth two rockets are launched each with a capable speed of 200,000 km/s. one goes "left" and the other goes "right".

in one second from earth frame, how far apart are the rockets?

8. Aug 25, 2004

### HallsofIvy

Get a good introductory book on relativity! 300000 was not just an "arbitrary number" (actually 300000 is not even correct). c is the calculated speed of light. The formulas for the relationship between speeds and different frames of reference, necessary in order that Maxwell's equations not be violated in different frames of reference, lead to the result that a speed can never be greater than c.

"the earth is one reference frame. from the earth two rockets are launched each with a capable speed of 200,000 km/s. one goes "left" and the other goes "right".

in one second from earth frame, how far apart are the rockets?"

The would, of course, be 400,000 km apart. But that does NOT say that they would be 400,000 km apart in the frame of one or the other of the rockets and so does NOT say what their speed RELATIVE TO ONE ANOTHER would be.

If one rocket has speed v1 relative to the earth and and the other, in exactly the opposite direction, has speed v2 relative to the earth, then their speed of each, relative to the other, is $\frac{v_1+v_2}{1+ \frac{v_1v_2}{c^2}}$.

In your example v1= v2= 200000 and c= 300000 so the speed of either rocket relative to the other is $\frac{400000}{1+ 0.44444..}$= 276923, still less than c.

9. Aug 25, 2004

your equation is wrong. There is no root in the denominator.

10. Aug 25, 2004

why are you using that equation when you know they're 400,000 km apart in one earth second?

seems to me the equation is just flat out wrong. and gets more wrong the more seconds go by.

and i have said before i know and understand the principles of maxwell's equations. i just don't see why we're needing to uphold them for relative speeds. in other words, what's so bad about them separating at 400,000 km/s ?

11. Aug 25, 2004

### JasonRox

How can you measure the relative speed of yourself with an object you can not see? Or is impossible to detect?

The light or any kind of communication from the rocket going left will NEVER get to the rocket going right. So how does either rocket calculate its speed relative to each other, without knowing where they are or even if they exist at all.

Let's say you are watch your buddy in the other rocket, 10 metres apart, and you plan on launching in the opposite directions, at the same time. Both of your rockets will reach speeds of 200 000km/h relative to EARTH, in an instant. I'm excluding acceleration because special relativity works for only constant velocities, so the ships go from 0 to 200 000km/h, in an instant. You both have a launch setup at 11:59:59pm, and at that very second you see his face, but when 12:00:00am hits your clock, he's gone. He doesn't even exist as far as you are concerned. You don't see the spaceship; you see NOTHING!

So, how do you measure your speed relative to something else you can't see. You can't even see a BLUR FOR CRYING OUT LOUD! Sure, you can say, on earth we can say that they are going 200,000km/h, but that's relative to us. You can't just do measurements for the rocket going to left, so he knows his speed relative to the other rocket. The measurements are relative to earth. You just can't change that.

I hope I'm thinking about it correctly. :)

12. Aug 25, 2004

thats just the conditions of Special Relativity. Its the way velocities are added in relativity, and its been proven mathamatically.

$$u' = \frac{u+v}{1+ \frac{uv}{c^2}}$$

13. Aug 25, 2004

### Alkatran

You're missing the point. A photon's speed depends on neither the source NOR the observer. This means that no matter how fast whatever the photon comes out of is going, it's speed is c. No matter how fast you're moving a photon's speed relative to you is c.

That is one of the basic things you need to understand and accept before relativity will make any sense. Those two spaceships seperating at 400 000 km/sec in the earth frame aren't going away from each other at c in THEIR frames.

14. Aug 25, 2004

If I'm understanding you correctly, you dont seem to understand. If you both travel at 200,000km/s in opposite directions, your relative velocity is NOT found by using a Gallilean transformation, you must use einsteins idea of special relativity. Here is how to solve the problem:

$$u' = \frac{u+v}{1+ \frac{uv}{c^2}}$$

$$u' = \frac{0.66c+0.66c}{1+ \frac{(0.66c)(0.66c)}{c^2}}$$

$$u' = \frac{1.33c}{1+ \frac{0.44c^2}{c^2}}$$

$$u' = \frac{1.33c}{1+ 0.44}$$

$$u' = \frac{1.33c}{1.44}$$

$$u' = \frac{1.33c}{1.44}$$

$$u' = 0.92c$$

The two friends see themselves movian apart at 0.92c. They would see eachother, since light is faster than this.

15. Aug 25, 2004

### Staff: Mentor

What makes you say that?
If you have the data (which we do) you can calculate the relative speed. For one rocket to actually measure the speed of the other would require some preparation. But so what?
No need for any "instant acceleration" fantasy. Just have the two rockets pass each other exactly at the point that the Earth clock reads 12:00:00am. What makes you think that one rocket can't see light from the other rocket?

If you know some physics, you can certainly calculate how fast one rocket would move as measured by the other. (HallsofIvy did that calculation several posts ago.) Don't get hung up on the practicalities of actually making the measurements. You'd need instruments, not just some guy staring out the window watching the other rocket race by. But there is nothing conceptually problematic about any frame measuring the speed of any object.

16. Aug 25, 2004

why should i care about measurements of distance speed and time FROM the spaceships? they're travelling at incredible speeds and therefore subjected to all manner of time dialations and length contractions. what should matter is the earth distances and the earth times we're using and therefore the relevant earth SPEED we derive from them.

what meaning does speed distance and time have if we don't first define a frame everyone accepts as THE BASIS for all measurements?

it's like trying to sell a hectare of land to someone but give everyone different definitions of the size of that unit of measurement. understanding is the key to communication after all :|

17. Aug 25, 2004

### Alkatran

The ships need their own measurements before they can convert them to earth-frame measurements. Also, the earth is rotating around the sun which is rotating the galaxy, the home-frame point of reference probably shouldn't be accelerating so much.

18. Aug 25, 2004

### Staff: Mentor

If you lived on that spaceship you might care. After all, wouldn't you want to know how far away that asteroid is and how soon it will smash into you?
We on earth are traveling just as much as they are. How do you like those crazy time dilations and length contractions?
If all you care about is Earth measurements, that's OK.

And what frame would you pick? Earth's? Why would a rocket dweller want to use Earth as a basis for measuring? He'd have to take all his direct measurements and do some wacky conversion. He won't be able to just look at his own watch, he'd have to keep asking "Now what would that Earth clock read?".
Actually, it's just the opposite. It's like you in the rocket are trying to sell me a boat: you say it's a 100 foot cruiser. But I, on earth, say: Nonsense, it's only a 4 foot dingy. I'll give you 10 bucks for it.

You seem to think that there is a frame out there that gives the "real" length. Not true: time, length, and synchronization really do depend on which frame is doing the measuring.

19. Aug 25, 2004

### HallsofIvy

Looks like a lot of people gave basically the same answer within a few minutes!

20. Aug 25, 2004

exactly what i was getting at. all "measurements" have no comparative "sense" unless we're functioning in the same frame. what's the point of measurements in the first place if we can't accurately use them to convey information?

why NOT use earth as a standard frame? we DID start off here after all. as far as a BASE is concerned it's probably the best one we could choose.

21. Aug 25, 2004

### Staff: Mentor

We know how to relate measurements made in different frames.
Who says we can't use measurements to convey information? But you'd better realize that measurement depend on the frame doing the measuring. To state a measurement without specifying the frame in which it was made is meaningless (where relativity is relevant).

For one, it would be a little silly and incredibly inconvenient for things moving at relativistic speeds. Imagine being in a rocket moving at some speed with respect to the earth. If the rocket inhabitants need to measure a distance or a time, doesn't it make sense for them to use their own perfectly good rulers and clocks? If, for some reason, we needed to know what an Earth observer would measure, we could do the conversion.

But get it straight: Different frames measure different distances and times. There is no escape.

22. Aug 26, 2004

okay, let's do a simple experiment.

i'm on a rocket heading towards earth at 200,000km/s. it just so happens that i'm 200,000 km from earth when a satellite 300,000 km from earth on the opposite side explodes. in the second it takes for that light from the explosion to reach earth, i travel 200,000 km and reach earth as well in time to see the explosion simultaneously as observers on earth.

taking my rocketship as a stationary frame, did the light NOT travel 500,000 km/s? the distance from my ship to the earth was 200,000km and the distance from the earth to the satellite was an additional 300,000km.

23. Aug 26, 2004

### JasonRox

If the rocket is considered stationary, then the satellite and the Earth is moving towards you, along with the exploded satellite.

It's kind of like driving on the highway. If you consider yourself stationary, the tree is flying past you, and the bird flying around the tree.

---------------------------------
My question about earlier. (Using the rockets going left and right.)

So when we say they are going 200 000km/s, what is really happenning to the "speedometer"? If the one rocket reads 200 000km/s, .66c, than the other spaceship is in fact only going (.92c - .66c) relative to them? What I am visualizing here, is that you are going this fast, but other objects going in the opposite direction seem or are going slower. If you are in rocket 1(200,000km/s), and you look into rocket 2 with a telescope to see their clock, you will find their clock ticks slower than yours. This is what is suppose to happen. Even though they might be going 200 000km/s, which you see with your scope on the "speedometer", together you not going above c. This is because one second has gone by for you (200 000km distance), but because their clock is SLOWER, a second hasn't passed for them yet, so they only travelled (.92c-.66c distance) relative to you.

I am pretty sure I understand it now. This makes so much more sense.

We can take it to earth now too. Sure one second has gone by for you, but if you look at the clock of rocket 1, a second did not go by at all.

Before I continue further on this, comment on this.

Last edited: Aug 26, 2004
24. Aug 26, 2004

### Staff: Mentor

OK.

Excellent. A perfectly unambiguous statement.
Says who? Once you start talking about distances and simultaneity, you must state what frame is doing the measuring. I will assume you mean that according to earth measurements you pass a point 200,000 km from earth just as a satellite that is 300,000 km from earth explodes. Note that these are observations made by earth observers. Other observers (such as you in the rocket) would measure different distances and would not agree that you passed that point at the same time that the satellite exploded.
Again, says who? I will assume you mean that according to earth observers, both you and the light from the explosion will reach the earth in one second.

No. Now that you are viewing things from a different frame there are several things to realize:
(1) The distance from you to the satellite is 500,000 km measured from earth. In your frame, that distance is only $L \sqrt{(1 - \frac{v^2}{c^2})} = L (0.74) = 370,000 km$. Similarly, you measure your distance to earth as only 148,000 km.
(2) Since you are 148,000 km from earth, your clocks tell you that it takes $t = D/v = 0.74$ seconds for you to reach earth.
(3) According to your frame, the satellite exploded before you reached the position in question. (Simultaneity is frame dependent: two events that appear to happen at the same time in one frame, will happen at different times in another frame.)​
Only when measured from the earth frame.

Last edited: Aug 26, 2004
25. Aug 26, 2004