Bounded by spacetime

1. Jul 7, 2008

epkid08

If I were to travel from point a to point b at the speed of light, given that a and b are real, I would travel a constant distance through space, and because space is bonded with time, traveling from point a to point b at the speed of light would result in a constant distance in time. It may seem as though time didn't affect me while I traveled, but literal time was taken.

Is this correct?

2. Jul 7, 2008

Mentz114

You can't, so don't worry about what you might or might not experience.

3. Jul 7, 2008

epkid08

Tell that to Einstein; why not just revert back to a traditional society with your logic...

4. Jul 7, 2008

cristo

Staff Emeritus
As Mentz says, the theory states that a massive body cannot travel at the speed of light. Thus, it is pointless speculating over such a question (and, as you well know, such speculation will not be allowed on PF).

5. Jul 7, 2008

HallsofIvy

Staff Emeritus
It's not necessary to tell it to Einstein- Einstein said the same thing as Mentz114.

It is, of course, possible for light to move at the speed of light. In reference to a photon, "It may seem as though time didn't affect me while I traveled, but literal time was taken" is correct. (except for the phrase "literal time", I have no idea what "literal" might mean there.) In the frame of reference of the photon, no time at all has passed. There will have been differing amounts of time measured in other inertial frames of reference, depending on their relative speeds, because, while the speed of light is the same in all, the distance between a and b will differ.

Notice the the words "constant distance" and "constant time" can only mean "constant" in a specific inertial frame of reference.

6. Jul 7, 2008

epkid08

Sorry, I'm just trying to understand this.
This is what I don't understand, if a photon traveled at the speed of light, it would appear as though, in both the reference frame of the photon, and an observer, that the photon wasn't affected by time at all. Why would traveling at the speed of light have this sort of affect? The speed of light is constant, so if a photon traveled at the speed of light over a distance of d, it would have taken literal time, t, t not being zero; T would be very very very very close to zero, but not zero. The frame of reference shouldn't matter whether or not the photon took time to travel that distance at the speed of light. Are you telling me that the photon traveling at the speed of light, from point a to point b, a and b being ANY number, took zero time in doing so?

7. Jul 7, 2008

Staff: Mentor

Photons don't have "reference frames" in the sense that we use them in relativity. If you try to use the Lorentz transformation equations to calculate what a photon "observes", starting from what you "observe" in your own reference frame, and using a relative speed of c, you end up trying to divide by zero, which is mathematically undefined.

("observe" = "determine something's space-time coordinates in the observer's reference frame")

8. Jul 7, 2008

DopplerDog

I think it makes sense to ask what happens in the limit as the traveller approaches the speed of light, even if the traveller can never reach it. Take two points, a and b, at rest relative to each other. Let's define A's frame as that stationary to both points a and b, and have A sit at a. Let's say traveller B is moving, relative to A, from a to b.

Relative to A, B's clock slows down. Relative to B, the distance from a to b contracts. This is in accordance with the Lorentz transformation.

As B approaches the speed of light, A sees B's clock slow down to almost nothing. B, on the other hand, measures the distance between points a and b as contracting to almost nothing.

Since there are no preferred frames, B sees A's clock slowing down to almost nothing.

What B experiences "at the speed of light" is meaningless for the reasons described in previous posts.

Last edited: Jul 7, 2008
9. Jul 7, 2008

epkid08

This is kinda what I was trying to get at, but isn't this saying that time naturally progresses at the speed of light?

10. Jul 7, 2008

DopplerDog

What do you mean by "time naturally progresses"?

Did you notice that the situation is symmetrical between A and B? A sees B's clock slow down, but B sees A's clock slow down. B doesn't experience anything substantially different from A, because there are no preferred frames: A's frame and B's frame are equivalent inertial frames (provided they are moving at uniform velocity to each other).

11. Jul 7, 2008

epkid08

Right now, we are traveling into the future. Right now, time is progressing at a speed. Traveling at the 'speed of time' would give a sense of frozen universe. Also, as you've stated, traveling at the speed of light does the same. That being said, time then progresses at the speed of light.

Anyways, has the Lorentz transformation been proven?

12. Jul 7, 2008

Peeter

You probably need to pick up an introductory book electromagnetism. Einstein also wrote a very readable layman's introduction to the relativity bits calls "Relativity, the Special and General theory". As a new learner myself I found that very palatable (it's probably available in your public library ... that's where I found it).

The way that I have found for myself that I like the best for an initial introduction of the Lorentz transformation is to look at the wave equation. Light appears as a wave regardless of your velocity, so if you take the wave equation for an electromagnetic field (ie: propagation of a signal at the speed of light) :

$$\partial_{xx} + \partial_{yy} + \partial_{zz} - \frac{1}{c^2}\partial_{tt}$$

, and perform a change of variables using the chain rule introducing a
along one direction x' = x - vt, then you don't get the wave equation anymore (you get a bunch of mixed terms too). This would imply a curious distortion of light signals with the velocity of the observer (ie: one that we don't observe). If you do the math for a linear change of variables in the above, then lo and behold, out falls the Lorentz transformation, as the linear transformation required to maintain the wave equation with respect to motion of the "observer".

You can do the same thing (with less math) by looking at invarience of the speed of a spherical light shell:

$$x^2 + y^2 +z^2 -c^2t^2 = {x'}^2 + {y'}^2 +{z'}^2 -c^2{t'}^2$$

and look for linear transformations that retain this form (ie: no mixed terms). My Berkeley physics mechanics book introduces the transform this way, but it was not all obvious to me that this was a sensical starting point when I first tried learning the subject (later I started to understand the somewhat subtle statement of what it really meant for the speed of light to be constant ... understanding or acceptance of that justifies the spherical shell equation above).

13. Jul 7, 2008

DopplerDog

Ok, I think I see the source of your confusion. It doesn't help to think of time as having a "speed". You could say a clock slows down relative to another, but the "speed of time" is not clearly defined.

Time is a coordinate, used to mark events. Think of your cartesian coordinates to mark a point in 3-D euclidean space: x, y, z. Time is an additional coordinate, except now you're not woking in 3-D, but in relativistic space-time (need 4 coordinates to specify an event). Clocks can be used to mark this coordinate. Special Relativity allows you to convert from one observer's coordinate system to another's.

The speed of light, however, is a fundamental constant of profound importance. What one observer sees as pure distance, another can see as a mixture of time and distance - and the speed of light provides the "conversion" factor between time and distance. I think this may be what you're getting at. Maybe.

We can say that the Lorentz transformation has been thoroughly tested, and a large number of experiments confirm its predictions to a high level of accuracy (experiments that could not otherwise be explained through Newtonian mechanics), e.g. subatomic particles have been measured to have longer lifetimes when travelling close to the speed of light.

Last edited: Jul 7, 2008
14. Jul 7, 2008

Mentz114

epkid08,

Physics is an empirical science, experiment and observation are the only ways to test if things are right. This link below is a 'must-read'.

I've just seen DopplerDog and Peeters posts. I agree completely ( if that's grammatical).

15. Jul 7, 2008

epkid08

Okay, I'll get on reading that, but one last question, shouldn't the force of gravity be a variable in the Lorentz Transformation?

16. Jul 7, 2008

DopplerDog

Well, the Lorentz Transformation assumes uniform motion in inertial frames (Special Relativity), and with gravity the issue is a little more complicated (requires an understanding of General Relativity).

However, I think what you're getting at is that gravity must somehow have a bearing in transforming beween coordinate systems, and if that's the case you'd be 100% right. It's just that the Lorentz Transformation (and Special Relativity) assumes inertial frames - you need to look at General Relativity to deal with gravity.

17. Jul 8, 2008

Mentz114

Yes. Except that SR is designed for no gravity. Einstein called his theory of gravity 'general' ( not special) relativity because it uses the same idea of 4D space-time as SR, but allows the space-time to be 'curved' by mass and energy.

18. Jul 8, 2008

yuiop

No, T is exactly zero.

Yes, zero time as measured by a hypothetical imaginary clock carried by the photon. A moving observer see all distances as length contracted by gamma. A hypothetical imaginary observer traveling at c would see all distances as zero. It is not surprising from that point of view that our hypothetical observer can travel from anywhere to anywhere in zero time by his own imaginary clock because he sees all distances as zero. The time t measured by any inertial observer with a real physical clock watching the photon would be t=d/c. This is an informal explanation as of course no sentient observer with rest mass can travel at c even in principle.

19. Jul 8, 2008

epkid08

I didn't realize that a photon had zero mass. My whole point was on a particle with mass.

20. Jul 9, 2008

cristo

Staff Emeritus
Well, as you have been told before, a massive body cannot travel at the speed of light. A photon has zero rest mass, and thus travels at the speed of light.