Time it will take when traveling at speed of light

welshvegan

I hope that you can bear with me on this, as I till cannot grasp:

1. Does length contraction only affect the astronaut and his spaceship?

2. How is 4 light years actually 1 light year if traveling at .99999c?

3. Why does time contract for the astronaut but not for a beam of light?

DaveC426913

It affects anything that is travelling relative to something else.

From Earth, the astronaut and his spaceship are travelling near c, and thus are length contracted.

From the astronaut's point of view, the entire universe is travelling with respect to him, and thus the entire universe is length contracted.


See above. When travelling near c, the astronaut measures Alpha Centauri as much closer than we would. It is also flattened.

Photons do not experience time.

welshvegan

1. If a photon does not experience time, then why does it take time to reach somewhere?

DaveC426913

Here's an analogy I came up with a while back.

Spacetime, that which everything including photons lives in, is 4-dimensional. Let's eliminate one of the dimensions (vertical) and put the time dimension in there instead. You now have cvomplete spatial freedom thpough the two horizontal spatial dimensions but you pass through the time dimension by moving vertically.

You are in a service elevator in a building under construction. The elevator ascends at one floor per second and it does not slow or stop or go backwards. This is you, moving through spacetime. Your spatial coordinates are not changing but your time coordinate is, as can be seen by the passing floors.

As you go up through level after level of floors, you realize that there are loooong strands of yarn tied to pillars from floor to floor at all sorts of angles. The first piece is tied to the the 2nd floor pillar right next to the elevator. The other end of that same piece is tied waaaaay up on the 6th floor to a pillar out on the north east corner of the building.

These strands are photons. They are joined at two ends, one end at point 2 in time (2nd floor), and the other at point 6 in time (6th floor).

You can only experience one floor at a time, so you only see sections of yarn that are horizontally in your line of sight. What you see is that the line of yarn (that is, only the short section you can see on the floor you're on) starts (is emitted) very close to you (right by the elevator) and "moves" away from you, ending up, four floors later, way out at the north east corner (where it is absorbed). When you were on the first floor there was no yarn to be seen, and when you passed the sixth floor, there was no yarn to be seen.

You have experienced this bit of yarn as an apparent movement through the building's space as a function of your travel through the floors (and, incidentally, through time). It was "emitted" on the second floor, and "absorbed" on the sixth floor.

But note that the yarn has experienced no travel at all through the vertical time dimension. It is static. It has had no experience of "moving" from floor to floor, no experience of emission or absorption - or of any "time" whatsoever.

Frame Dragger

There was a thread here a month or two ago about that subject, and I don't believe a consensus was reached.

Fredrik

1. No. The length contraction formula tells you how to calculate the length of anything in one coordinate system, given its length in another coordinate system. In other words: The coordinate system that we associate with Alpha Centauri's motion assigns a length to the spaceship that's shorter by a factor of gamma than the length assigned by the coordinate system associated with the spaceship's motion. And the coordinate system that we associate with the spaceship's motion assigns a distance between the ship and Alpha Centauri that's shorter by a factor of gamma than the distance assigned by the coordinate system associated with Alpha Centauri's motion.

2. There is no "actual" distance between the sun and Alpha Centauri. The motion of these stars traces out two curves (called "world lines") in spacetime. Their small velocities relative to each other are negligible, so we can treat the world lines as parallel. The "distance" between the sun and Alpha Centauri would have to be the length of some curve in spacetime that connects a point on one of the world lines with a point on the other. But which two points should we choose, and which curve should we connect them with? The "obvious" choice is to choose two points that are simultaneous and the shortest possible curve between them. But different observers will disagree about which point on AC's world line is simultaneous with a given point on the sun's world line. So one observer will say that the distance between the stars is the length of a specific curve connecting their world lines, and another observer will say that the distance is the length of some other curve.

3. See this quote and the thread I linked to in there:
It doesn't. Not in its own rest frame, because it doesn't have a rest frame. (Note that this doesn't mean that it experiences zero time. It just means that the concept of "experienced time" is undefined).

The reason why it takes time for a photon to get somewhere is of course that its speed is finite. Speed is defined as distance/time, so 0 time means infinite speed.

There are threads about it every month, but have you ever heard a reasonable argument for why it should make sense to define a photon's point of view by doing a Lorentz transformation with speed v and taking the limit v→c? (I don't consider "Brian Greene did it" an actual argument. )

welshvegan

Thanks guys so much, it is very slowly getting clearer. I'm going to continue to do a lot more studying on this, as it is no way as simple as I thought. In our worldly everyday idea of time, it is just very difficult for me to get my brain around the concept of someone aging slower at .99999c than someone on earth, and how a light year is not a light year for someone traveling at .99999c, but is a light year for a beam of light. Whew! My brain is gonna explode!!!!

Fredrik

A light year for you isn't a light year for the guy on the rocket, but there's no such thing as "for a beam of light", so the last statement doesn't make sense.

I recommend that you study spacetime diagrams. That's the best way to learn about SR. (It's almost impossible to understand SR without them, and they are much easier to understand than the algebraic expressions representing Lorentz transformations).

welshvegan

There is such thing as a beam of light. If I shine a light from a flashlight, that is a beam of light, right? If not, then what is it? It also takes time to reach somewhere.

I will study the the spacetime diagrams. Thanks :)

Frame Dragger

Can't argue with that... and "Brian Greene did it" has never passed these lips! lol.

Fredrik

Right. There's nothing wrong with the phrase "a beam of light". The phrase that doesn't make sense is "for a beam of light". Massless particles don't have experiences. Massive particles don't either of course, but when we're dealing with a massive particle, there's at least a natural way to associate a coordinate system with its motion. Not so for massless particles, as I explained. So when we're dealing with a massive object, we can talk about its "experiences" even if it's incapable of having experiences. We define its "point of view" to be the description of events in terms of the coordinate system we associate with its motion.

If we can't associate a coordinate system with a particle's motion, we can't give meaning to a question like "How much time passes for such a particle?", or to any question or statement about anything for that particle.

Yes, in the coordinate system associated with your motion, and in the coordinate systems associated with the motion of other massive objects, but not in a coordinate system associated with its motion, because there's no such thing.