# Photons and dimensions

1. Dec 24, 2003

### terwilligerjones

OK...I'm a certifiable idiot, and an old one at that, but this has nagged at me for decades and perhaps someone here could explain it to me:

If I were to accelerate towards the speed of light, I would perceive other, non-accelerating, frames of reference as foreshortened in the direction of my travel, right?

If I were traveling with a beam of light, then, would I not perceive the universe of my travel as a plane perpendicular to my "direction" of travel? And if I did, could I have any notion of motion, reflection, refraction, direction or any of the myriad other qualities that are observable about a beam of light? Stated another way, would not I perceive the beginning and end of my journey to be simultaneous with every point through which I passed, none of which I could perceive?

That just doesn't sound right....there shouldn't be one frame of reference (mine) where no motion is detectable, and others where it is.

Sorry if that's a dumb one, but...

terwilliger

2. Dec 24, 2003

### NateTG

In general Special Relativity is about velocity, and General Relativity is about acceleration.

The fundemental assumption of Special Relativity is that the Speed of Light is constant in all reference frames. This means that traveling 'with' a photon is impossible. Reference frames where the observer is traveling at c are invalid.

3. Dec 24, 2003

### StarThrower

Let us suppose that F1 is an inertial reference frame. In fact, let us define a rectangular coordinate system in F1, whose origin is the center of mass of the universe (CMU). Now, let the axes of this coordinate system not be spinning with respect to the stars. Suppose that at the beginning of your journey, you are at rest in F1. Let |v| denote your speed in F1. Now by Newton's first law an object at rest will remain at rest, unless acted upon by an outside force. Now you turn on your engines, and there is a force on your ship, and your ship begins to accelerate in F1, according to Newton's second law.
Eventually you turn off your rocket engines, and you now are just coasting at a constant speed |v| in F1. Now, let L0 denote the length of your ship in F1 at the beginning of your journey (when you were at rest in F1). According to the theory of relativity, your length in F1 is now less than L0. The formula for your current length L in reference frame F1 is:

$$L = L_0 \sqrt{1-v^2/c^2}$$

Now, let us suppose that there is a clock located at the CMU, which ticks at regular intervals. Suppose that an identical clock was on your ship at the beginning of your journey (when you were at rest), thus at the beginning of your journey, your clock ticked at the same rate as the CMU clock.

Now, according to the theory of relativity, your clock no longer ticks at the same rate as the CMU clock, it ticks slower. The formula which relates an amount of time of an event in the CMU frame, to the amount of time for the same event according to the ships clock is:

$$\Delta t = \frac{\Delta t'}{\sqrt{1-v^2/c^2}}$$

So suppose that the units of the CMU clock are seconds, and that the time of an event in the CMU frame is exactly one second. The time of the same event in the ships frame can be found from the following equation:

$$1 = \frac{\Delta t'}{\sqrt{1-v^2/c^2}}$$

Thus, an event which takes one second in the CMU frame now takes:

$$\Delta t' = \sqrt{1-v^2/c^2}$$

So for example, suppose that delta t' is 1/4 of a second. It follows that your ship accelerated to a final speed |v| given by:

1/16 = 1-v^2/c^2 ---> v^2/c^2 = 1 -1/16 = 15/16

---> v^2 = 15c^2/16

which implies that

$$v = \sqrt{15c^2/16}$$

Which as you can see is less than the speed of light c=299792458 meters per second.

So the point is, that the rate at which your ships clock ticks, is no longer the same as the rate at which the CMU clock ticks, and in fact your clock is ticking slower, the CMU clock is ticking faster. If you were to reach the speed of light, your clock would stop ticking, and so according to the theory of relativity, no ship can be accelerated from rest to the speed of light.

Ok so we begin with all of that. Now your question is kind of vague. You are asking about other reference frames being shortened. Here is how I understand you (correct me if I am wrong later).

Suppose that you started out at rest, and then accelerated towards our sun. Once you turned off your engines, you were at rest in an inertial frame F2. Now in F2, the sun is coming at you with speed |v|. And so I think your question is about the distance which the sun is away from you. You are asserting that it has length contracted in frame F2. So that if the distance from the sun to the origin of F2 is 1000*299792458 meters in a "sun at rest" system, that the distance the sun is away from you in frame F2 is less than 1000*299792458 meters, because of distance contraction. (you are asking yes or no)

Well, suppose that the distance you are away from the sun the moment you turn off your engines in your system is also 1000*299792458 meters. Now consider a photon fired from the sun at you. In the suns frame, a clock there will tick out 1000 seconds as the time it takes the photon to reach you. But, your clock is supposed to tick slower than a clock at rest with respect to the sun. Thus, the time it takes the photon to reach you should end up being less, and so you won't come up with the same speed of light, it will travel the same distance it traveled in the suns frame, in less time, and so you will think the speed of light is greater than c in your frame. According to relativity, this is not possible, and so the distance from you to the sun in your frame must be contracted, so that in your frame that distance is less than 1000*299792458 meters. So the answer to your question is yes, if the theory of relativity is correct.

It isn't a dumb question, but it sure is a confusing one. Certainly if clocks in your frame don't tick, then there is no passage of time in that frame, and so hence no motion is detectable, and yes that is very strange.

I think your first question is more interesting though, about the length contraction.

4. Dec 24, 2003

### StarThrower

No nate. The postulate of the special theory of relativity, is that the speed of light in any inertial system is c. There is nothing in relativity which forbids you from looking at things in a frame which is traveling along with a photon. However, if the special theory is correct, then a frame in which a photon is at rest is a non-inertial reference frame, and you shouldn't expect Newton's laws to be satisfied in such a frame. But to simply say such a frame is 'invalid' isn't the same as saying such a frame is 'non-inertial'. The latter is correct, not the former.

5. Dec 24, 2003

### NateTG

Ok, then how do you (non trivially) reconcile these two statements:

1. The photon is traveling at c in this reference frame.
(This is a fundemental assumption of SR)
2. The photon is has a speed of 0 in this reference frame.
(This is synonymous with 'this reference frame is at rest w.r.t. the photon.)

The trivial reconcilliation, BTW, would be that c=0, but I think you can accept that that doesn't work.

6. Dec 24, 2003

### StarThrower

I don't understand your question nate.

Suppose that a baseball is at rest in some inertial reference frame. Set up a three dimensional rectangular coordinate system, with the origin at the center of mass of the baseball. Now, suppose that a photon moves past the baseball, by the postulate of relativity, the speed of this photon with respect to the center of mass of the baseball is c. Now, there is nothing whatsoever wrong with analyzing the motion of the baseball in a coordinate system whose origin is the photon.

In this frame, the photon never strays from the origin (hence the photon is at rest in this frame), but rather it is the baseball which is moving. Now, because the theory of relativity asserts that the speed of light is c in any inertial reference frame, and the speed of light in this reference frame is clearly not c (it is 0) it follows that if the postulate of relativity is correct, then this frame is not an inertial reference frame.

Here is what I am saying:

Let V denote the speed of the photon in the baseball's frame

V = c

Let V denote the speed of the photon in the photon frame

V = 0

7. Dec 24, 2003

### NateTG

You're right.

I'm not being sufficiently clear in my thought and expression. I meant invalid in the sense that physics does not make predictions about the reference frame, only about approaching it.

I missed the important distinction between inertial and non-inertial reference frames.

8. Dec 27, 2003

### terwilligerjones

I was a bit vague. I suppose my original question had to do with observations made perpendicular to the direction of travel.

Suppose I am moving with constant velocity in your F1 frame of reference centered at the CMU. My velocity in that frame of reference is 35 mph and as I pass our solar system, I make detailed measurements. As I fly by the earth, I find that the earth has a mean diameter of d, and a mean distance from the sun of one a.u. As I fly by the the sun, I measure its mean diameter as D.

Later, I retrace this same journey at a constant velocity of (15/16)c. It's my understanding that my measurements on this trip will result in values in accordance with the Lorentz-Fitgerald contraction equations you cited. The earth's mean diameter will be less. The mean distance from the sun and the sun's mean diameter will appear smaller in my second set of measurements. I think.

This is where I get confused. Can you help me with this?

terwilliger

9. Dec 28, 2003

### TillEulenspiegel

I think I grasp what your inquiring.
The state of your vehicle @35MPH does have some length contraction in the direction of travel. It is not really measurable as the quantity is so small, but can be expressed mathematically.

The contraction at near light speed is more pronounced not only is there time dilation but again length contraction as well. That holds true for Your ship, You and everything on board. So, if you took a yardstick at that first pass and used to make a arbitrary measurement of d of the earth and sun and marked them on the yardstick the contractions at 35mph is trivial and will not affect this measurement. The second pass when you use that same yardstick...is more problematic because at 15/16c the contraction will make the yardstick stretched, so when You make the second measurement it APPEARES to You that the Earth and Sun have shrunk, when in fact you and your measuring device have changed and not the bodies you measure.In fact it dosn't have to be a yardstick at all all of your instrumentalities are effected like the clock =).
I hope that's clear.

10. Dec 29, 2003

### terwilligerjones

Well, we're closer, at any rate. Suppose I were to take my ship as the origin of a frame of reference, F2. In that frame of reference, the solar system would first slide by me at 35 mph. I would make my measurements of the mean distance to the sun from the earth, the earth's diameter and the sun's diameter. Later, the solar system would flash by me at 15/16c, and I could repeat the measurements.

Would not my results in the second measurement cause me to conclude that the diameters and mean distance were now smaller by sqr(1-(v^2/c^2))? [?]

terwilliger

11. Dec 29, 2003

### NateTG

That depends on whether you account for relativistic contraction. Otherwise, you would also come to the conclusion that the sun had changed in color from red to Ultraviolet, and that the motion of the planets sponateously slowed down as you passed.

12. Jan 1, 2004

### terwilligerjones

From my frame of reference, as the solar system slides by me at greater and greater velocity, is it true that as the velocity increases, my observations of the mean orbital distance between earth and sun, and the diameters of earth and sun, observed out a window perpendicular to the direction of travel, will approach zero?

I think this is where I get stuck, as if that is true, it would seem that as one approaches a light speed differencial between the observer and the solar system, the appearance of the solar system from the observer's frame of reference would approach a plane perpendicular to the direction of travel.

13. Jan 1, 2004

### TillEulenspiegel

"the appearance of the solar system from the observer's frame of reference would approach a plane perpendicular to the direction of travel."
I'm not sure what your trying to say. The whole point of Einstien's take was that all inertial planes of reference are relative to any other frame. Read that to mean that they are basically arbitrary, there is no plane of absolute non-motion ( I.E.) no yardstick that works identically , all the time,for all planes except the speed of light. You could say that Your ship and the solar system were like elevators one going up , the other down ( and vise versa ) or past each other like 2 ships, or if measured by another plane ( like Andromeda ) maybe a parabola. All these observations are based on which plane the observer is in. You still possess a vector of travel but again , it is arbitrary.