# Special Relativity Contains Massive Error

1. Dec 17, 2003

### Tempest

Suppose an atom that is at rest in an inertial reference frame simultaneously emits two identical photons in exactly the opposite direction. For all intent and purpose consider the radius of the atom as vanishingly small. Now consider things after one second has passed in this 'atomic frame'.

According to the theory of SR, any photon in an inertial reference frame must have speed c=299792458 meters per second in that frame. Thus, after one second has passed, the distance each photon is away from the atom, in the atomic frame absolutely has to be 299792458 meters. Let us denote this distance by D.

Now consider things from the point of view of a coordinate system whose origin is one of the photons. In this system, the speed of this photon is zero, and the atom is moving (lets say to the left) at speed c, and the other photon is moving to the left at a speed greater than c.

Now, either the distance the atom is away from the photon in this photonic frame has been Lorentz contracted or not. Assume it has been contracted. Let d denote the distance the atom and photon are separated by in the photonic frame, after 1 second has passed in the atomic frame.

The exact amount of contraction is given by:

d = D(1-v^2/c^2)^1/2 = D(1-c^2/c^2)^1/2 = 0

Hence, after one second has passed in the atomic frame, the distance between the two photons in the photonic frame is zero. Hence, in the photonic frame, the two photons are not in relative motion, which is impossible (that would only be true if they were emitted in the same direction, and they were emitted in opposite directions). Thus, the Galilean transformations hold when we switch from the atomic frame to the photonic frame.

Now, consider a third reference frame F3, which was initially moving away from the atom (to the right) at a constant speed S, and such that after one second had passed in the atomic frame, photon 1 was located at the origin of F3.

We know that in the atomic frame, that after one second has passed photon one traveled 299792458 meters. Suppose that S = 1 meter per second. Thus, F3 is moving away from the atom at a constant speed of one meter per second.

And after one second has passed in the atomic frame, we have stipulated that the photon is at the origin of F3. Therefore, at the moment the photon's were emitted, the origin of F3 was located 299792457 meters away from the atom, in the atom's frame. It would then follow that after one second had passed in the atomic frame, that the origin of F3 would be 299792458 meters away from the atom, and hence the photon would be at the origin of F3 as stipulated, so everything is fine.

Hence in the atomic frame (atom at rest), we have an event which begins when the origin of F3 is 299792457 meters away from the atom, and ends when the origin of F3 is 299792458 meters away from the atom, and this event takes 1 second in the atomic frame.

(work in progress)

2. Dec 17, 2003

### Staff: Mentor

If the atom thinks 1 second has passed, how much time does each photon think has passed?

3. Dec 17, 2003

### chroot

Staff Emeritus
Perhaps you should have titled this thread: "DOES SR contain a massive error?"

The answer is that, no, it does not.
You've chosen to use a coordinate system which is singular. In other words, for a photon, time effectively does not exist. The distance between itself and any other point in the universe is length-contracted to zero, and the time it takes to get anywhere in the universe is also zero.

In other words, in the photon's rest frame, the atom, the other photon, and your Aunt Sally are all at the same distance: zero. And they always will be. Thus, there is no paradox, or error, or whatever you'd like to call it.

I can understand that it can be a little challenging to learn and intuitively understand special relativity -- but you need to think a little more before trying to convince us that it's wrong.

I should also note that this sort of discussion is permitted on physicsforums.com only in the Theory Development forum (which is a subforum of the General Physics forum).

- Warren

4. Dec 17, 2003

### chroot

Staff Emeritus
A very pertinent question...

- Warren

5. Dec 17, 2003

### Tempest

Re: Re: Special Relativity Contains Massive Error

6. Dec 17, 2003

### Tempest

7. Dec 17, 2003

### chroot

Staff Emeritus
Re: Re: Re: Special Relativity Contains Massive Error

What I mean is that if you were the captain of a hypothetical starship capable of traveling at c (it's hypothetical, of course, because a starship necessarily has mass), you would be able to travel to any point in the universe in exactly zero time. This means that every point in the universe appears to be zero distance from you.

In more mathematical terms, length contraction can be expressed as:

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

where $L_0$ is the distance between two objects when you're not moving at all relative to them, and $L$ is their distance when you are travelling at some velocity [itex]v[/tex]. As you can see,

$$\lim_{v \rightarrow c} L = 0$$

In other words, when you're going c, everything in the universe is zero distance from you.

- Warren

8. Dec 17, 2003

### chroot

Staff Emeritus
That would be an incorrect answer, Tempest. The answer is zero. Do you know how to use the Lorentz transformations to demonstrate time dilation?

- Warren

9. Dec 17, 2003

### Tempest

It was shown that the galilean transformations hold when swtiching from the atomic frame to the photonic frame, since the event took one second in the atomic frame, it also took one second in the photonic frame. Do you mean derive the time dilation formula from the Lorentz coordinate transformations? Probably I do.

Last edited by a moderator: Dec 17, 2003
10. Dec 17, 2003

### chroot

Staff Emeritus
The Galilean transformation never holds. It is a low-velocity approximation of the Lorentz transformation.

Surely, if you're going to be dealing with things going at the speed of light, you're no longer able to use a low-velocity approximation.

- Warren

11. Dec 17, 2003

### Tempest

If the galilean transformation never holds, then the distance the two photons are separated after one second has passed in the atomic frame is zero, hence the two photons are not in relative motion, hence the two photons are traveling in the same direction, contrary to the stipulation that they were emitted in opposite directions. Hence, the galilean transformation sometimes holds.

Last edited by a moderator: Dec 17, 2003
12. Dec 17, 2003

### chroot

Staff Emeritus
Nope. Sorry. The very concept of 'relative motion' really just has no meaning for photons, since they don't experience distance (or time).

And no, the Galilean transformation never, ever holds. Special relativity replaces it with a generalization, the Lorentz transform.

You're welcome to say "Special relativity has a massive error if I assume the Galilean transformation holds," but that's not very useful -- since the Galilean transformation never holds in special relativity.

- Warren

13. Dec 17, 2003

### Staff: Mentor

Hey, I'm not completely useless outside the engineering forums.

Tempest, the very title of your thread should set off a blinking red light in the back of your mind when you read it back to yourself. If SR did contain a massive error, don't you think it would have been noticed before? The implications of what you are saying are so fundamental (it really is a massive error you are suggesting) that the theory would fall apart completely and all the technology we use that depends on it (GPS for example) would not work.

You need to check to see if your warning light is working properly.

Last edited: Dec 17, 2003
14. Dec 18, 2003

### HallsofIvy

Doesn't it occur to you that those two statements are contradictory? The only way you can assert the second is to deny the first. In other words your whole argument is "If special relativity is wrong, then it contains an error"!

15. Dec 18, 2003

### Tempest

Naaaah.

GPS would still work, because it does work AND SR is wrong.

16. Dec 18, 2003

### Tempest

I) IF the speed of light is c in any inertial reference frame then the photonic frame isn't an inertial reference frame.

II) If in some inertial reference frames the speed of light isn't c then the photonic frame is an inertial reference frame.

So if SR doesn't contain a massive error then the photonic frame is a non-inertial reference frame.

Hence, if the photonic frame is an inertial reference frame, then SR contains a massive error. :)

Last edited by a moderator: Dec 18, 2003
17. Dec 18, 2003

### chroot

Staff Emeritus
Correct logic, but the premise is invalid -- a frame centered on a photon is not inertial.

- Warren

18. Dec 18, 2003

### Tempest

How sure are you?

19. Dec 18, 2003

### chroot

Staff Emeritus
110%. Do you understand the definition of an inertial frame?

- Warren

20. Dec 18, 2003

### Tempest

Refresh my memory, something about a frame which isn't accelerating if I do recall. Only problem with that, is acceleration is relative.

I am thinking more along the lines of a frame in which all three of Newton's laws are true.

Last edited by a moderator: Dec 18, 2003